Permutation And Combination
Name: _____________________
Date: _____________________
Instructions: Answer all questions. Write your answers clearly in the space provided.
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
20 men handshake with each other without repetition. What is the total number of handshakes made?
In how many different ways can the letters of the word OPERATE be arranged ?
Out of 5 women and 4 men, a committee of three members is to be formed in such a way that at least one member is a women. In how many different ways can it be done ?
In how many ways a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women ?
In how many different ways can the letters of the word GAMBLE be arranged?
In how many different ways can the letters of the word RUMOUR be arranged?
In how many different ways can the letters of the word ‘TRANSPIRATION’ be arranged so that the vowels always come together?
In how many different ways can the letters of the word CAPITAL be arranged so that the vowels always come together?
A committee of 5 members is to be formed by selecting out of 4 men and 5 women. In how many different ways the committee can be formed if it should have 2 men and 3 women?
In how many different ways can the letters of the word INCREASE be arranged?
In how many different ways can the letters of the word CREATE be arranged?
A college has 10 basketball players. A 5 member's team and a captain will be selected out of these 10 players. How many different selections can be made?
How many four letter distinct initials can be formed using the alphabets of English language such that the last of the four words is always a consonant?
How many number of times will the digit 7 be written when listing the integers from 1 to 1000?
In how many ways can 15 people be seated around two round tables with seating capacities of 7 and 8 people?
In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION ?
A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?
A tea expert claims that he can easily find out whether milk or tea leaves were added first to water just by tasting the cup of tea. In order to check this claims 10 cups of tea are prepared, 5 in one way and 5 in other. Find the different possible ways of presenting these 10 cups to the expert.
A committee is to be formed comprising 7 members such that there is a simple majority of men and at least 1 woman. The shortlist consists of 9 men and 6 women. In how many ways can this committee be formed?
How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?
In how many different ways can the letters of the word ENGINEERING be arranged?
In how many different ways can the letters of the word CORPORATION be arranged so that the vowels may occupy only the odd positions?
In how many different ways can the letters of the word JUDGE be arranged in such a way that the vowels always come together?
In how many different ways can the letters of the word DISPLAY be arranged?
In how many different ways can the letters of the word RIDDLED be arranged?
In how many different way can the letters of the word WEDDING be arranged?
A select group of 4 is to be formed from 8 men and 6 women in such a way that the group must have at least 1 women. In how many different ways can it be done ?
In how many different ways can the letters of the word AUCTION be arranged in such a way that the vowels always come together?
In how many different ways can the letters of the word MACHINE be arranged so that the vowels may occupy only the odd positions?
In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?
How many Permutations of the letters of the word APPLE are there?
How many different words can be formed using all the letters of the word ALLAHABAD? (a) When vowels occupy the even positions. (b) Both L do not occur together.
In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.
A two member committee comprising of one male and one female member is to be constitute out of five males and three females. Amongst the females. Ms. A refuses to be a member of the committee in which Mr. B is taken as the member. In how many different ways can the committee be constituted ?
In how many ways 2 students can be chosen from the class of 20 students?
Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?
A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.
Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side.
A man positioned at the origin of the coordinate system. the man can take steps of unit measure in the direction North, East, West or South. Find the number of ways of he can reach the point (5,6), covering the shortest possible distance.
There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?
a, b, c, d and e are five natural numbers. Find the number of ordered sets (a, b, c, d, e) possible such that a + b + c + d + e = 64.
There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when each of the 5 numbers is divided by 4 is written down on another card (the 6th card) in order. How many sequences can be written down on the 6 th card?
From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee are possible?
Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
Jay wants to buy a total of 100 plants using exactly a sum of Rs. 1000. He can buy Rose plants at Rs. 20 per plant or marigold or Sun flower plants at Rs. 5 and Rs. 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?
How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?
In how many ways can a committee of 4 people be chosen out of 8 people?
In how many different ways can the letters of the word EXTRA be arranged so that the vowels are never together?
In how many different ways can the letters of the word ‘BAKERY’ be arranged?
In how many different ways can the letters of the word DAILY be arranged?
In how many different ways can letters of the word OFFICES be arranged?
In how many different ways can the letters the word FORMULATE be arranged?
In an examination there are three multiple choice questions and each question has 4 choices. The number of ways in which a student can fail to get all answer correct is-
In how many different ways can the letters of the word AWARE be arranged?
A committee of 5 members is to be formed out of 3 trainees, 4 professors and 6 research associate. In how many different ways can this be done if the committee should have 2 trainees and 3 research associates?
In how many different ways can the letters of the word CREAM be arranged?
In how many different ways can the letters of the word ALLAHABAD be arranged?
From a group of 7 men 6 women, 5 persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
In how many different ways can the letters of the word BANKING be arranged in such a way that the vowels always come together?
A box contains 2 white, 3 black and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least 1 black ball is to be included in the draw?
Out of 5 men and 3 women, a committee of three members is to be formed so that it has 1 women, and 2 men. In how many different ways can it be done?
In how many different ways can the letters of the word BANANA be arranged?
A committee of 5 members is to be formed out of 3 trainees, 4 professors and 6 research associates. In how many different ways can this be done, if the committee should have 4 professors and 1 research associate or all 3 trainees and 2 professors?
How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?
How many different five-letter words can be formed using the letter from the world APPLE?
I have an amount of Rs. 10 lakh, which I went to invest in stocks of some companies. I always invest only amounts that are multiples of Rs 1 lakh in the stock of any company. If I can choose from among the stocks of five different companies, In how many ways can I invest the entire amount that I have?
A selection committee is to be chosen consisting of 5 ex-technicians. Now there are 12 representatives from four zones. It has further been decided that if Mr. X is selected, Y and Z will not be selected and vice-versa. In how many ways it can be done?
How many 5-digit positive integers exist the sum of whose digits are odd?
There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?
In how many ways can seven friends be seated in a row having 35 seats, such that no two friends occupy adjacent seats?
How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters?
In a hockey championship, there are 153 matches played. Every two team played one match with each other. The number of teams participating in the championship is:
A box contains 10 balls out of which 3 are red and rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same colour?
A book-shelf can accommodate 6 books from left to right. If 10 identical books on each of the languages A,B,C and D are available, In how many ways can the book shelf be filled such that book on the same languages are not put adjacently.
In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number:
In how many ways can 3 men and their wives be made stand in a line such that none of the 3 men stand in a position that is ahead of his wife?
In a cricket match if a batsman score 0, 1, 2, 3, 4 or 6 runs of a ball, then find the number or different sequences in which he can score exactly 30 runs of an over. Assume that an over consists of only 6 balls and there were no extra and no run outs.
In how many ways can the letters of the word MANAGEMENT be rearranged so that the two As do not appear together?
How many numbers are there between 100 and 1000 such that at least one of their digits is 6?
A student is required to answer 6 out of 10 questions divided into two groups each containing 5 questions. He is not permitted to attempt more than 4 from each group. In how many ways can he make the choice?
While packing for a business trip Mr. Debashis has packed 3 pairs of shoes, 4 pants, 3 half-pants,6 shirts, 3 sweater and 2 jackets. The outfit is defined as consisting of a pair of shoes, a choice of "lower wear" (either a pant or a half-pant), a choice of "upper wear" (it could be a shirt or a sweater or both) and finally he may or may not choose to wear a jacket. How many different outfits are possible?
How many positive integers 'n' can be form using the digits 3, 4, 4, 5, 6, 6, 7 if we want 'n' to exceed 60,00,000?
In how many ways can the letters of the word ‘MOMENT’ be arranged?
There are six teachers. Out of them two are primary teachers and two are secondary teachers. They are to stand in a row, so as the primary teachers, middle teachers and secondary teachers are always in a set . The number of ways in which they can do so, is-
In how many different ways can the letters of the word SOFTWARE be arranged in such a way that the vowels always come together?
A committee of 5 members is to be formed by selecting out of 4 men and 5 women. In how many different ways the committee can be formed if it should have at least 1 man?
In how many ways can the letters of the word MATHEMATICS be arranged so that all the vowels always come together?
In how many different ways can the letters of the word TOTAL be arranged?
In how many different ways can the letters of the word SMART be arranged?
$$left( {{}^{75}{P_2} - {}^{75}{C_2}}
ight) = ?$$
In how many different ways can the letters of the word ABSENTEE be arranged?
In how many different ways can the letters of the word OPERATE be arranged?
A local delivery company has three packages to deliver to three different homes. if the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?
How many natural numbers less than a lakh can be formed with the digits 0,6 and 9?
There are 20 couples in a party. Every person greets every person except his or her spouse. People of the same sex shake hands and those of opposite sex greet each other with a Namaste (It means bringing one's own palms together and raising them to the chest level). What is the total number of handshakes and Namaste's in the party?
In how many ways can a leap year have 53 Sundays?
In a certain laboratory, chemicals are identified by a colour-coding system. There are 20 different chemicals. Each one is coded with either a single colour or a unique two-colour pair. If the order of colours in the pairs does not matter, what is the minimum number of different colours needed to code all 20 chemicals with either a single colour or a unique pair of colours?
In how many ways can 6 green toys and 6 red toys be arranged, such that 2 particular red toys are never together whereas 2 particular green toys are always together?
There are five comics numbered from 1 to 5. In how many ways can they be arranged, so that part-1 and part-3 are never together?
The number of ways which a mixed double tennis game can be arranged amongst 9 married couples if no husband and wife play in the same is:
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?
There are five women and six men in a group. From this group a committee of 4 is to be chosen. How many different ways can a committee be formed that contain three women and one man?
Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct.
In a railway compartment, there are 2 rows of seats facing each other with accommodation for 5 in each, 4 wish to sit facing forward and 3 facing towards the rear while 3 others are indifferent. In how many ways can the 10 passengers be seated?
There are three prizes to be distributed among five students. If no students gets more than one prize, then this can be done in:
A teacher of 6 students takes 2 of his students at a time to a zoo as often as he can, without taking the same pair of children together more than once. How many times does the teacher go to the zoo?
A family consist of a grandfather, 5 sons and daughter and 8 grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the 4 seats at each end and the grandfather refuses to have a grandchild on either side of him. The number of ways in which the family can be made to sit is:
If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?
What is the value of 1 × 1! + 2 × 2! + 3 × 3! + . . . . . . . . n × n! where n! means n factorial or n(n-1) (n-2) . . . . . . . . 1
When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?
A question paper consists of three sections 4,5 and 6 questions respectively. Attempting one question from each section is compulsory but a candidate need not attempt all the questions. In how many ways can a candidate attempt the questions?
In how many ways a President, VP and Water-boy can be selected from a group of 10 people.
In an examination paper, there are two groups each containing 4 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected?
After every get-together every person present shakes the hand of every other person. If there were 105 handshakes in all, how many persons were present in the party?
How many diagonals can be drawn in a pentagon?
The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?
A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged?
If $$5{ imes ^{ ext{n}}}{{ ext{P}}_3} = 4{ imes ^{left( {{ ext{n}} + 1}
ight)}}{{ ext{P}}_{3,}}$$ xa0 xa0find n?
Ten different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, the number of words which have at least one letter repeated is:
12 chairs are arranged in a row and are numbered 1 to 12. 4 men have to be seated in these chairs so that the chairs numbered 1 to 8 should be occupied and no two men occupy adjacent chairs. Find the number of ways the task can be done.
How many words can be formed by re-arranging the letters of the word ASCENT such that A and T occupy the first and last position respectively?
If 6 P r = 360 and If 6 C r = 15, find r ?
In how many ways can six different rings be worn on four fingers of one hand?
There are 7 non-collinear points. How many triangles can be drawn by joining these points?
From 6 men and 4 ladies, a committee of 5 is to be formed. In how many ways can this be done, if the committee is to include at least one lady?
The number of ways of arranging n students in a row such that no two boys sit together and no two girls sit together is m(m > 100). If one more student is added, then number of ways of arranging as above increases by 200%. The value of n is:
How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4 if repetition of digits is allowed?
How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated.
There are 10 seats around a circular table. If 8 men and 2 women have to seated around a circular table, such that no two women have to be separated by at least one man. If P and Q denote the respective number of ways of seating these people around a table when seats are numbered and unnumbered, then P : Q equals
How many factors of 2 5 × 3 6 × 5 2 are perfect squares?
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
In how many ways can the letters of the word 'LEADER' be arranged?
There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices.
In a party every person shakes hands with every other person. If there are 105 hands shakes, find the number of person in the party.
The number of positive integers which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 without repetition.
In the next World cup of cricket there will be 12 teams, divided equally in two groups. Teams of each group will play a match against each other. From each group 3 top teams will qualify for the next round. In this round each team will play against each others once. Four top teams of this round will qualify for the semifinal round, where they play the best of three matches. The Minimum number of matches in the next World cup will be:
There are 10 person among whom two are brother. The total number of ways in which these persons can be seated around a round table so that exactly one person sit between the brothers , is equal to:
If letters of the work KUBER are written in all possible orders and arranged as in a dictionary, then the rank of the word KUBER will be:
A letter lock consists of 4 rings, each ring contains 9 non-zero digits. This lock can be opened by setting four digit code with the proper combination of each of the 4 rings. Maximum how many codes can be formed to open the lock ?
10 students are to be seated in two rows equally for the Mock test in a room. There are two sets of papers, Code A and Code B. each of two rows can have only one set of paper but different that from other row. In how many ways these students can be arranged ?
How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?
Find the number of ways in which 8064 can be resolved as the product of two factors?
Six boxes are numbered 1, 2, 3, 4, 5 and 6. Each box must contain either a white ball or a black ball. At least one box must contain a black ball and boxes containing black balls must be consecutively numbered. find the total number of ways of placing the balls.
In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex pentagons of distinctly different areas can be drawn using these points as vertices?
What is the total number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girlfriends such that each of them gets at least one gift?
Answer Key
ight) + left( {^6{C_2}{ imes ^4}{C_2}}
ight) + $$ xa0 xa0 xa0$$left( {^6{C_3}{ imes ^4}{C_1}}
ight) + $$ xa0 $$left( {^6{C_4}}
ight)$$ $$ = left( {^6{C_1}{ imes ^4}{C_1}}
ight) + left( {^6{C_2}{ imes ^4}{C_2}}
ight) + $$ xa0 xa0 xa0$$left( {^6{C_3}{ imes ^4}{C_1}}
ight) + $$ xa0 $$left( {^6{C_2}}
ight)$$ $$ = left( {6 imes 4}
ight) + left( {frac{{6 imes 5}}{{2 imes 1}} imes frac{{4 imes 3}}{{2 imes 1}}}
ight) + $$ xa0 xa0 xa0 $$left( {frac{{6 imes 5 imes 4}}{{3 imes 2 imes 1}} imes 4}
ight) + $$ xa0xa0 $$left( {frac{{6 imes 5}}{{2 imes 1}}}
ight)$$ $$eqalign{
& = left( {24 + 90 + 80 + 15}
ight) cr
& = 209 cr} $$
& { ext{Required}},{ ext{number}},{ ext{of}},{ ext{ways}} cr
& = {{}^8{C_5} imes {}^{10}{C_6}} cr
& = {{}^8{C_3} imes {}^{10}{C_4}} cr
& = {frac{{8 imes 7 imes 6}}{{3 imes 2 imes 1}} imes frac{{10 imes 9 imes 8 imes 7}}{{4 imes 3 imes 2 imes 1}}} cr
& = 11760 cr} $$
& = left( {{}^3{C_1} imes {}^6{C_2}}
ight) + left( {{}^3{C_2} imes {}^6{C_1}}
ight) + left( {{}^3{C_3}}
ight) cr
& = left( {3 imes frac{{6 imes 5}}{{2 imes 1}}}
ight) + left( {frac{{3 imes 2}}{{2 imes 1}} imes 6}
ight) + 1 cr
& = left( {45 + 18 + 1}
ight) cr
& = 64 cr} $$
& { ext{Required number of ways}} cr
& = {{}^7{C_5} imes {}^3{C_2}} cr
& = {{}^7{C_2} imes {}^3{C_1}} cr
& = {frac{{7 imes 6}}{{2 imes 1}} imes 3} cr
& = 63 cr} $$
ight)left( {2!}
ight)}}$$ xa0 = 10080 Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = $$frac{{4!}}{{2!}}$$ = 12 ∴ Required number of words = (10080 x 12) = 120960
& = frac{{7!}}{{2!}} cr
& = frac{{7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{2} cr
& = 2520 cr} $$
ight) + left( {{}^5{{ ext{C}}_2} imes {}^4{{ ext{C}}_1}}
ight)$$ xa0 xa0 $$ + left( {{}^5{{ ext{C}}_3}}
ight)$$ $$ = left( {5 imes frac{{4 imes 3}}{{2 imes 1}}}
ight)$$ xa0 $$ + left( {frac{{5 imes 4}}{{2 imes 1}} imes 4}
ight)$$ xa0 $$ + left( {frac{{5 imes 4 imes 3}}{{3 imes 2 imes 1}}}
ight)$$ $$eqalign{
& = left( {30 + 40 + 10}
ight) cr
& = 80 cr} $$
& left( {{}^8{C_5} imes {}^{10}{C_6}}
ight) + left( {{}^8{C_3} imes {}^{10}{C_4}}
ight) cr
& = frac{{8 imes 7 imes 6}}{{3!}} imes frac{{10 imes 9 imes 8 imes 7}}{{4!}} cr
& = frac{{8 imes 7 imes 6}}{{3 imes 2 imes 1}} imes frac{{10 imes 9 imes 8 imes 7}}{{4 imes 3 imes 2 imes 1}} cr
& = 11760 cr} $$
& {}^6{P_6} = 6! cr
& = {6 imes 5 imes 4 imes 3 imes 2 imes 1} cr
& = 720 cr} $$
& = frac{{6!}}{{2! imes 2!}} cr
& = frac{{6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{2 imes 2}} cr
& = 180 cr} $$
& = frac{{9! imes 5!}}{{2!, 2! ,2! ,2! ,2!}} cr
& = frac{{9 imes 8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 5 imes 4 imes 3 imes 2}}{{2 imes 2 imes 2 imes 2 imes 2}} cr
& = 1360800 cr} $$
& = {{}^4{C_2} imes {}^5{C_3}} cr
& = {{}^4{C_2} imes {}^5{C_2}} cr
& = {frac{{4 imes 3}}{{2 imes 1}} imes frac{{5 imes 4}}{{2 imes 1}}} cr
& = 60 cr} $$
& = frac{{8!}}{{2!}} cr
& = frac{{8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{2} cr
& = 20160 cr} $$
& = frac{{6!}}{{2!}} cr
& = frac{{6 imes 5 imes 4 imes 3 imes 2!}}{{2!}} cr
& = 360 cr} $$
Now, the captain can be selected from these 6 players in 6 ways. Therefore, total ways the selection can be made is 210 × 6 = 1260 Alternatively We can select the 5 member team out of the 10 in 10 C 5 ways = 252 ways.
The captain can be selected from amongst the remaining 5 players in 5 ways.
Therefore, total ways the selection of 5 players and a captain can be made, = 252 × 5 = 1260
The first three letters can be either consonants or vowels. So, each of them have 26 options. Note that the question asks you to find out the number of distinct initials and not initials where the letters are distinct.
Hence, required answer = 26 × 26 × 26 × 21 = 26 3 × 21
If arranging these 'n' objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.
i.e., number of arrangements = $$frac{{left( {n - 1}
ight)!}}{2}$$
You can choose the 7 people to sit in the first table in 15 C 7 ways.
After selecting 7 people for the table that can seat 7 people, they can be seated in:
(7 - 1)! = 6!
The remaining 8 people can be made to sit around the second circular table in:
(8 - 1)! = 7! Ways.
Hence, total number of ways: 15 C 7 × 6! × 7!
The vowels occupy 3 rd , 5 th , 7 th and 8 th position in the word and the remaining 5 positions are occupied by consonants.
As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the before mentioned 4 places and the consonants can occupy 1 st , 2 nd , 4 th , 6 th and 9 th positions.
The 4 vowels can be arranged in the 3 rd , 5 th , 7 th and 8 th position in 4! Ways.
Similarly, the 5 consonants can be arranged in 1 st , 2 nd , 4 th , 6 th and 9 th
We may divide the 8 students as follows: Case I: 5 students in the first car and 3 in the second.
Hence, 8 students are divided into groups of 5 and 3 in 8 C 3 = 56 ways. Case II: 4 students in the first car and 4 in the second. So, 8 students are divided into two groups of 4 and 4 in 8 C 4 = 78 ways. Therefore, the total number of ways in which 8 students can travel is: 56 + 70 = 126
& = frac{{10!}}{{5! imes 5!}} cr
& = 252 cr} $$
Let the number of required alphabets in the language be 'n'. Therefore, using 'n' alphabets we can form n × n × n = n 3 distinct 3 digit initials. NOTE: Distinct initials are different from initials where the digits are different.
For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different. This n 3 different initials = 1 million. i.e. n 3 = 10 6 (1 million =10 6 ) n 3 = 10 2 3 n = 10 2 n = 100
Hence, the language needs to have a minimum of 100 alphabets
& = frac{{11!}}{{3! ,3! ,2! ,2! ,1!}} cr
& = frac{{11 imes 10 imes 9 imes 8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{6 imes 6 imes 2 imes 2 imes 1}} cr
& = left( {11 imes 10 imes 9 imes 8 imes 7 imes 5}
ight) cr
& = 277200 cr} $$
& = frac{{7!}}{{2!}} cr
& = frac{{7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{2 imes 1}} cr
& = 2520 cr} $$ Now, (OOAIO) has 5 letters, out of which we have 3O, 1A and 1I. Number of ways of arranging these letters $$eqalign{
& = frac{{5!}}{{3!}} cr
& = frac{{5 imes 4 imes 3 imes 2 imes 1}}{{3 imes 2 imes 1}} cr
& = 20 cr} $$ ∴ Required number of ways = (2520 × 20) = 50400
& {}^7{P_7} = 7! cr
& = left( {7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}
ight) cr
& = 5040 cr} $$
& = frac{{7!}}{{3!}} cr
& = frac{{7 imes 6 imes 5 imes 4 imes 3!}}{{3!}} cr
& = left( {7 imes 6 imes 5 imes 4}
ight) cr
& = 840 cr} $$
& = frac{{7!}}{{2!}} cr
& = frac{{7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{2 imes 1}} cr
& = 2520 cr} $$
ight) + left( {{}^6{C_2} imes {}^8{C_2}}
ight)$$ xa0 xa0 $$ + left( {{}^6{C_3} imes {}^8{C_1}}
ight)$$ xa0 $$ + left( {{}^6{C_4} imes {}^8{C_0}}
ight)$$ $$ = left{ {6 imes frac{{8 imes 7 imes 6}}{{3 imes 2 imes 1}}}
ight} + $$ xa0xa0 $$left( {frac{{6 imes 5}}{{2 imes 1}} imes frac{{8 imes 7}}{{2 imes 1}}}
ight)$$ xa0 $$ + left( {frac{{6 imes 5 imes 4}}{{3 imes 2 imes 1}} imes 8}
ight)$$ xa0xa0 $$ + left( {{}^6{C_2} imes 1}
ight)$$ $$ = left{ {6 imes frac{{8 imes 7 imes 6}}{{3 imes 2 imes 1}}}
ight}$$ xa0xa0 $$ +, 420, + $$ xa0$$left( {frac{{6 imes 5 imes 4}}{6} imes 8}
ight)$$ xa0 $$ + left( {frac{{6 imes 5}}{{2 imes 1}} imes 1}
ight)$$ $$ = left( {336 + 420 + 160 + 15}
ight)$$ $$ = 931$$
ight)left( {mathop {}limits^2 }
ight)left( {mathop {}limits^3 }
ight)left( {mathop {}limits^4 }
ight)left( {mathop {}limits^5 }
ight)left( {mathop {}limits^6 }
ight)left( {mathop {}limits^7 }
ight)$$ Now, 3 vowels can placed at any of the three places out of four marked 1, 3, 5, 7 Number of ways of arranging the vowels $$eqalign{
& = {}^4{P_3} cr
& = left( {4 imes 3 imes 2}
ight) cr
& = 24 cr} $$ 4 consonants at the remaining 4 positions may be arranged in $${}^4{P_4} = 4! = $$ xa0
24 ways Required number of ways = (24 × 24) = 576
& = frac{{5!}}{{2!}} cr
& = frac{{120}}{2} cr
& = 60 cr} $$
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers. These 9 papers can be arranged in 9! Ways. And two papers can be arranged themselves in 2! Ways. No. of arrangement when best and worst paper do not come together, = 10! - 9! × 2! = 9!(10 - 2) = 8 × 9!
& { = ^{20}}{{ ext{C}}_2} cr
& = frac{{20!}}{{2! imes 18!}} cr
& = 20 imes frac{{19}}{2} cr
& = 190 cr} $$
& { = ^6}{{ ext{C}}_2} cr
& = frac{{6!}}{{2! imes 4!}} cr
& = 15 cr} $$
& { = ^{10}}{{ ext{C}}_8}{ imes ^{10}}{{ ext{C}}_5} cr
& = frac{{10!}}{{2! imes 8!}} imes frac{{10!}}{{5! imes 5}} cr
& = left{ {10 imes frac{9}{2}}
ight} imes left{ {frac{{10 imes 9 imes 8 imes 7 imes 6}}{{5 imes 4 imes 3 imes 2 imes 1}}}
ight} cr
& = 11340 cr} $$
& { = ^8}{{ ext{C}}_3} cr
& = frac{{8 imes 7 imes 6}}{{1 imes 2 imes 3}} cr
& = 56 cr} $$
= 3 C 1 × 4! × 4! = 1728
The number of ways in which these steps can be taken is given by: = $$frac{{11!}}{{5! imes 6!}}$$ = 462
If arranging these n objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number. i.e., number of arrangements = $$frac{{left( {n - 1}
ight)!}}{2}$$ Let there be exactly one person between the two brothers as stated in the question. If we consider the two brothers and the person in between the brothers as a block, then there will 17 others and this block of three people to be arranged around a circle. The number of ways of arranging 18 objects around a circle is in 17! ways. Now the brothers can be arranged on either side of the person who is in between the brothers in 2! ways. The person who sits in between the two brothers could be any of the 18 in the group and can be selected in 18 ways. Therefore, the total number of ways = 18 × 17! × 2 = 2 × 18!
i.e., o, o, o, o, o, o . . . . (64 balls).
We have 63 gaps where we can place a wall in each gap, since we need 5 compartments we need to place only 4 walls. We can do this in 63 C 4 ways.
Now we are left with (6 - 1) men and (4 - 2) ladies (Mrs. X is not willing to join). We can choose 1 more in 5+2 C 1 = 7 ways. case (ii): If Mr. Y is not a member then we left with (6 + 4 - 1) people.
we can select 3 from 9 in 9 C 3 = 84 ways. Thus, total number of ways is 7 + 84 = 91 ways.
Let number of marigold plants be b .
Let the number of Sunflower plants be c .
According to question, 20a + 5b + 1c = 1000 - - - - - - (1) a + b + c = 100 - - - - - - - - - - (2) Solving the above two equations by eliminating c,
19a + 4b = 900 b = $$frac{{900 - 19a}}{4}$$ xa0 = $$225 - frac{{19a}}{4}$$ xa0 - - - - - - - (3) b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e . 0 < b < 99 - - - - - - - (4) Substituting (3) in (4), 0 < 225 - $$frac{{19a}}{4}$$ < 99 ⇒ 225 < -$$frac{{19a}}{4}$$ < (99 - 225) ⇒ 4 × 225 > 19a > 126 × 4 ⇒ $$frac{{900}}{{19}}$$ > a > 504 a is the integer between 47 and 27 - - - - - - - - (5) From (3), it is clear, a should be multiple of 4. Hence, possible values of a are (28,32,36,40,44) For a=28 and 32, a+b>100 For all other values of a, we get the desired solution: a=36,b=54,c=10 a=40,b=35,c=25 a=44,b=16,c=40 Three solutions are possible.
3 vowels can be selected in 4 C 3 ways.
Therefore, total number of groups each containing 4 consonants and 3 vowels, = 12 C 4 × 4 C 3 Each group contains 7 letters, which can be arranging in 7! ways.
Therefore required number of words, = 12 C 4 × 4 C 3 × 7!
& = {}^8{C_4} cr
& = frac{{8 imes 7 imes 6 imes 5}}{{4 imes 3 imes 2 imes 1}} cr
& = 70 cr} $$
& = frac{{7!}}{{2!}} cr
& = frac{{7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{2 imes 1}} cr
& = 2520 cr} $$
& = {}^9{P_9} cr
& = 9! cr
& = left( {9 imes 8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}
ight) cr
& = 362880 cr} $$
& = frac{{5!}}{{2!}} cr
& = frac{{5 imes 4 imes 3 imes 2 imes 1}}{2} cr
& = 60 cr} $$
& = left( {{}^3{C_2} imes {}^6{C_3}}
ight) cr
& = left( {{}^3{C_1} imes {}^6{C_3}}
ight) cr
& = left( {3 imes frac{{6 imes 5 imes 4}}{{3 imes 2 imes 1}}}
ight) cr
& = 60 cr} $$
& = 5! cr
& = left( {5 imes 4 imes 3 imes 2 imes 1}
ight) cr
& = 120 cr} $$
& = frac{{9!}}{{4! ,2! ,1! ,1! ,1!}} cr
& = frac{{9 imes 8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{4 imes 3 imes 2 imes 1 imes 2}} cr
& = 7560 cr} $$
ight) + $$ xa0 $$left( {{}^7{C_4} imes {}^6{C_1}}
ight) + $$ xa0 $$left( {{}^7{C_5} imes {}^6{C_0}}
ight)$$ $$ = left{ {frac{{7 imes 6 imes 5}}{{3!}} imes frac{{6 imes 5}}{{2!}}}
ight}$$ xa0 xa0 $$ + left( {{}^7{C_3} imes {}^6{C_1}}
ight)$$ xa0 $$ + left( {{}^7{C_2} imes 1}
ight)$$ $$ = left{ {frac{{7 imes 6 imes 5}}{6} imes frac{{6 imes 5}}{{2 imes 1}}}
ight}$$ xa0 xa0 $$ + left( {frac{{7 imes 6 imes 5}}{{3 imes 2 imes 1}} imes 6}
ight)$$ xa0xa0 $$ + left( {frac{{7 imes 6}}{{2 imes 1}} imes 1}
ight)$$ $$eqalign{
& = left( {525 + 210 + 21}
ight) cr
& = 756 cr} $$
& left( {{}^3{C_1} imes {}^6{C_2}}
ight) + left( {{}^3{C_2} imes {}^6{C_1}}
ight) + left( {{}^3{C_3}}
ight) cr
& = left{ {3 imes frac{{6 imes 5}}{{2 imes 1}}}
ight} + left( {frac{{3 imes 2}}{{2 imes 1}} imes 6}
ight) + 1 cr
& = left( {45 + 18 + 1}
ight) cr
& = 64 cr} $$
& = left( {{}^3{C_1} imes {}^5{C_2}}
ight) cr
& = 3 imes frac{{5 imes 4}}{{2 imes 1}} cr
& = 30 cr} $$
& = frac{{6!}}{{3! imes 2!}} cr
& = frac{{6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{6 imes 2}} cr
& = 60 cr} $$
ight) + $$ xa0 $$left( {{}^3{C_3} imes {}^4{C_{2}}}
ight)$$ $$eqalign{
& = left( {1 imes 6}
ight) + left( {1 imes frac{{4 imes 3}}{2}}
ight) cr
& = left( {6 + 6}
ight) cr
& = 12 cr} $$
ight) left(2!
ight) left(1!
ight) left(1!
ight)}}$$ = 60
Out of these 90000 positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number. Hence, there are $$frac{{90000}}{2}$$xa0 = 45000, 5-digit positive integers whose sum add up to an odd number.
The subsequent places can be filled in 3 ways each. Hence, the number of ways = 4 × 3 × 3 × 3 × 3 × 3 = 4 × 3 5
If the 3 vowels have to appear together as stated in the question, then there will 3 consonants and a set of 3 vowels grouped together.
One group of 3 vowels and 3 consonants are essentially 4 elements to be rearranged. The number of possible rearrangements is 4! The group of 3 vowels contains two a s and one u The 3 vowels can rearrange amongst themselves in $$frac{{3!}}{{2!}}$$ ways as the vowel a appears twice. Hence, the total number of rearrangements in which the vowels appear together are: $$frac{{4! imes 3!}}{{2!}}$$
The problem requires us to find out the number of outcomes in which the two As do not appear together.
The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter. Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in: $$frac{{9!}}{{2! imes 2! imes 2!}}$$ xa0 ways. Therefore, the required answer in which the two As do not appear next to each other
Unit digit could take any value of the 9 values (0 to 9, except 6) Tens Digit could take any value of the 9 values (0 to 9, except 6) Hundreds digit could take any value of the 8 values (1 to 9, except 6) numbers between 100 and 1000 which have at least one digit as 6, = 900 - 648 = 252
Number of ways of attempting more than 4 from a group, = 2 × 5 C 5 × 5 C 1 = 10
Required number of ways = 210 - 10 = 200
= 3 C 1 = 3 ways
Number of ways lower wear can be selected = (3 + 4) = 7 ways.
Number of ways upper wear can be selected,
= 3 + 6 + (3 × 6) = 27 ways
Number of ways jacket can be chosen or not chosen = 3 ways {No jacket, 1 st jacket, 2 nd jacket}. Total number of different outfits = 3 × 7 × 27 × 3 = 1701 ways
If the first digit is 6, the other digits can be arranged in $$frac{{6!}}{{2!}}$$ = 360 ways.
If the first digit is 7, the other digits can be arranged in $$frac{{6!}}{{2! imes 2!}}$$ xa0= 180 ways. Thus required possibilities for n, = 360 + 180 = 540 ways
& = frac{{6!}}{{2!}} cr
& = frac{{6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{2 imes 1}} cr
& = 360 cr} $$
ight) + left( {{}^4{C_2} imes {}^5{C_3}}
ight)$$ xa0 xa0xa0 $$ + left( {{}^4{C_3} imes {}^5{C_2}}
ight)$$ xa0 $$ + left( {{}^4{C_6} imes {}^5{C_1}}
ight)$$ $$ = left( {{}^4{C_1} imes {}^5{C_1}}
ight) + left( {{}^4{C_2} imes {}^5{C_2}}
ight)$$ xa0 xa0xa0 $$ + left( {{}^4{C_1} imes {}^5{C_2}}
ight)$$ xa0 $$ + left( {{}^4{C_4} imes {}^5{C_1}}
ight)$$ $$ = left( {4 imes 5}
ight) + left( {frac{{4 imes 3}}{{2 imes 1}} imes frac{{5 imes 4}}{{2 imes 1}}}
ight)$$ xa0 xa0xa0 $$ + left( {4 imes frac{{5 imes 4}}{{2 imes 1}}}
ight)$$ xa0 $$ + left( {1 imes 5}
ight)$$ $$ = left( {20 + 60 + 40 + 5}
ight)$$ $$ = 125$$
& = frac{{8!}}{{2!.2!}} cr
& = frac{{8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{2 imes 1 imes 2 imes 1}} cr
& = 10080 cr} $$ Now, (AEAI) has 4 letters, out of which we have 2A, 1E and 1I. Number of ways of arranging these letters $$eqalign{
& = frac{{4!}}{{2!}} cr
& = frac{{4 imes 3 imes 2 imes 1}}{2} cr
& = 12 cr} $$ ∴ Required number of ways = (10080 × 12) = 120960
& = frac{{5!}}{{2!}} cr
& = frac{{5 imes 4 imes 3 imes 2!}}{{2!}} cr
& = 60 cr} $$
& = {}^5{P_5} cr
& = 5! cr
& = left( {5 imes 4 imes 3 imes 2 imes 1}
ight) cr
& = 120 cr} $$
& = left( {{}^{75}{P_2} - {}^{75}{C_2}}
ight) cr
& = left{ {frac{{75!}}{{75! - 2!}} - frac{{75 imes 74}}{2}}
ight} cr
& = frac{{75!}}{{73!}} - left( {75 imes 37}
ight) cr
& = frac{{75 imes 74 imes 73!}}{{73!}} - left( {75 imes 37}
ight) cr
& = left( {75 imes 74 - 75 imes 37}
ight) cr
& = 75 imes 37 imes left( {2 - 1}
ight) cr
& = left( {75 imes 37}
ight) cr
& = 2775 cr} $$
& = frac{{8!}}{{3!}} cr
& = frac{{8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{6} cr
& = 6720 cr} $$
& = frac{{8!}}{{2!}} cr
& = frac{{8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{2} cr
& = 20160 cr} $$
One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.
We can calculate each of these cases and then add them together, or approach this problem from a different angle. The only case which is left out of the condition is the case where no wrong packages are delivered. If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above. There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package. The first person must get the correct package and the second person must get the correct package and the third person must get the correct package. ⇒ 1 × 1 × 1 = 1 Determine the total number of ways the three packages can be delivered. ⇒ 3 × 2 × 1 = 6 The number of ways at least one house gets the wrong package is: ⇒ 6 - 1 = 5 Therefore there are 5 ways for at least one house to get the wrong package.
The required numbers are from 1 to 99999
The numbers are five digit numbers.
Therefore, every place can be filled by 0, 6 and 9 in 3 ways.
Total number of ways = 3 × 3 × 3 × 3 × 3 = 3 5 But 00000 is also a number formed and has to be excluded.
Total number of numbers,
= 3 5 - 1 = 243 - 1 = 242
When a man meets a woman, there are two Namastes, whereas when a man meets a man (or a woman) there is only 1 handshake.
Number of handshakes, = 2 × 20 C 2 (men and women ) = 2 × 190 = 380 For a number of Namastes, Every man does 19 Namastes (to the 20 women excluding his wife) and they respond in the same way.
Number of Namastes = 2 × 20 × 19 = 760
Total number of Namastes and Handshake = 760 + 380 = 1140
So to have 53 Sundays one of these two days must be a Sunday.
This can occur in only 2 ways.
i.e. (Saturday and Sunday) or (Sunday and Monday). Thus number of ways = 2
We can arrange the 6 green toys and the remaining 4 red toys (excluding the 2 who are not to be together) is:
= 9! × 2! × 10 C 2 × 2! = 18 × 10!
The total number of ways in which part-1 and part-3 are always together:
= 4! × 2! = 48
Therefore, the total number of arrangements, in which they are not together is:
= 120 - 48 = 72
We know that in a mixed double match there are two males and two females. Step I: Two male members can be selected in 9 C 2 = 36 ways. Step II: Having selected two male members, 2 female members can be selected in, 7 C 2 = 21 ways. Step III: Two male and two female members can arranged in a particular game in 2 ways. Total number of arrangements = 36 × 21 × 2 = 1512 ways
When 10 coins are tossed simultaneously, the total number of outcomes = 2 10 Out of these, if the third coin has to turn up a head, then the number of possibilities for the third coin is only 1 as the outcome is fixed as head. Therefore, the remaining 9 coins can turn up either a head or a tail = 2 9
Let's sort out what we have and what we want. Have: 5 women, 6 men. Want: 3 women AND 1 man. The word AND means multiply. Woman and Men $$eqalign{
& ^{{ ext{have}}}{{ ext{C}}_{{ ext{want}}}}{ imes ^{{ ext{have}}}}{{ ext{C}}_{{ ext{want}}}} cr
& { = ^5}{{ ext{C}}_3}{ imes ^6}{{ ext{C}}_1} cr
& = 60 cr} $$
Therefore, the teacher goes to the zoo 15 times.
= 1 grandfather + 5 sons and daughters + 8 grandchildren = 14 The grandchildren can occupy the 4 seats on either side of the table in 4! = 24 ways. The grandfather can occupy a seat in (5 - 1) = 4 ways (4 gaps between 5 sons and daughter). And, the remaining seats can be occupied in 5! = 120 ways (5 seat for sons and daughter). Hence total number of required ways, = 8! × 480
In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it. So far 4 question there are 2 × 2 × 2 × 2 ways to attempt. As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 2 4 - 1
Similarly for the 2 nd section there are 2 5 - 1 ways in which he can attempt and for the 3 rd section there are 2 6 - 1 ways. The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections. Thus, total number of ways in which he can attempt questions in that paper: = (2 4 - 1)(2 5 - 1)(2 6 - 1) = 15 × 31 × 63 = 29295
Thus, total ways of selecting a President, VP and Water-boy from a group of 10 people would be 10 P 3
2 question from first group and 3 question from second group
Or
3 question from first group and 2 question from second group.
= ( 4 C 2 × 4 C 3 ) + ( 3 C 4 × 4 C 2 ) = 24 + 24 = 48
Then, $$eqalign{
& frac{{x imes left( {x - 1}
ight)}}{2} = 105 cr
& x = 15 cr} $$
Now, with A in the beginning, the remaining letters can be permuted in 5! ways. Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways. With L in the beginning, the first word will be LABORU, the second will be LABOUR. Hence, the rank of the word LABOUR is, 5! + 5! + 2 = 120 + 120 + 2 = 242 Note: 5! = 5 × 4 × 3 × 2 × 1 = 120
ight)!}}$$
There are 7 non-collinear points.
The number of triangles formed, $$eqalign{
& { = ^7}{{ ext{C}}_3} cr
& = frac{{7 imes left( {7 - 1}
ight) imes left( {7 - 2}
ight)}}{{3!}} cr
& = frac{{7 imes 6 imes 5}}{{3 imes 2 imes 1}} cr
& = 7 imes 5 cr
& = 35 cr} $$
⇒ n = 2a Then the number of arrangements = 2 × a! × a! If one more students is added, then number of arrangements, = a! × (a + 1)! But this is 200% more than the earlier ⇒ 3 × (2 × a! × a!) = a! × (a + 1)! ⇒ a + 1 = 6 and a = 5 ⇒ n = 10 But if n is odd, then number of arrangements,
= a!(a + 1)! Where, n = 2a + 1 When one student is included, number of arrangements, = 2(a + 1)! (a + 1)!
By the given condition, 2(a + 1) = 3, which is not possible.
The total number of cases of arranging 8 men and 2 women, so that women are together,
⇒ 8! ×2!
The number of cases where in the women are not together,
⇒ 9! - (8! × 2!) = Q
Now, when the seats are numbered, it can be considered to a linear arrangement and the number of ways of arranging the group such that no two women are together is,
⇒ 10! - (9! × 2!)
But the arrangements where in the women occupy the first and the tenth chairs are not favorable as when the chairs which are assumed to be arranged in a row are arranged in a circle, the two women would be sitting next to each other.
The number of ways the women can occupy the first and the tenth position,
= 8! × 2!
The value of P = 10! - (9! × 2!) - (8! × 2!) Thus P : Q = 10 : 1
& = left( {{}^7{C_3} imes {}^6{C_2}}
ight) + left( {{}^7{C_4} imes {}^6{C_1}}
ight) + left( {{}^7{C_5}}
ight) cr
& = left( {frac{{7 imes 6 imes 5}}{{3 imes 2 imes 1}} imes frac{{6 imes 5}}{{2 imes 1}}}
ight) + left( {{}^7{C_3} imes {}^6{C_1}}
ight) + left( {{}^7{C_2}}
ight) cr
& = 525 + left( {frac{{7 imes 6 imes 5}}{{3 imes 2 imes 1}} imes 6}
ight) + left( {frac{{7 imes 6}}{{2 imes 1}}}
ight) cr
& = left( {525 + 210 + 21}
ight) cr
& = 756 cr} $$
& = left( {{}^7{C_3} imes {}^4{C_2}}
ight) cr
& = left( {frac{{7 imes 6 imes 5}}{{3 imes 2 imes 1}} imes frac{{4 imes 3}}{{2 imes 1}}}
ight) cr
& = 210 cr} $$ Number of groups, each having 3 consonants and 2 vowels = 210 Each group contains 5 letters. Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120 ∴ Required number of ways = (210 x 120) = 25200
$$ = frac{{6!}}{{left( {1!}
ight)left( {2!}
ight)left( {1!}
ight)left( {1!}
ight)left( {1!}
ight)}} = 360$$
Thus, required number of triangle can be formed, = 10 C 3 - 4 C 3 = 120 - 4 = 116
$$eqalign{
& ^n{{ ext{C}}_2} = 105 cr
& { ext{or, }}frac{{n!}}{{2! imes left( {n - 2}
ight)!}} = 105 cr
& { ext{or, }}frac{{n imes left( {n - 1}
ight)}}{2} = 105 cr
& { ext{or, }}{n^2} - n = 210 cr
& { ext{or, }}{n^2} - n - 210 = 0 cr
& { ext{or, }}n = 15,, - 14 cr} $$
But, we cannot take negative value of n
So, n = 15 i.e. number of persons in the party = 15
Two digit positive numbers = 25
Three digit positive numbers = 100 4 digit positive numbers = 300 5 digit positive numbers = 600 Six digit positive numbers = 600 Total positive numbers, = 5 + 25 + 100 + 300 + 600 + 600 = 1630
So, required number of ways,
= 9 × 9 × 9 × 9
= 9 4
Remaining 5 will be seated in, = 5 C 5 ways
Students of each row can be arranged as,
= 5! × 5! ways
Two sets of paper can be arranged themselves in,
= 2! ways
Thus, Total arrangement,
= 10 C 5 × 5! × 5! × 2
= 7257600
B 2 will receive prize from rest 3 available prizes(so 3 ways)
B 3 will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways Hence, the 4 prizes can be distributed in 24 ways
2,3
3,4
4,5 and 5,6) and so on.
If there are 6 black balls, they can be placed in 1 way.
The total number of ways of placing the balls is = 1 + 2 + 3 + 4 + 5 + 6 = 21
The boxes are identical. If none of the boxes is to remain empty, then we can pack the toys in one of the following ways: Case i. 2, 2, 1 Case ii. 3, 1, 1 Case i: Number of ways of achieving the first option 2, 2, 1 Two toys out of the 5 can be selected in 5 C 2 ways. Another 2 out of the remaining 3 can be selected in 3 C 2 ways and the last toy can be selected in 1 C 1 way. However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2. Therefore, total number of ways of achieving the 2, 2, 1 option is: $$frac{{^5{C_2}{ imes ^3}{C_2}}}{2} = frac{{10 imes 3}}{2} = 15,{ ext{ways}}$$ Case ii: Number of ways of achieving the second option 3, 1, 1 Three toys out of the 5 can be selected in $$^5{C_3}$$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way. Therefore, total number of ways of getting the 3, 1, 1 option is $$^5{C_3}$$ = 10 ways. Total ways in which the 5 toys can be packed in 3 identical boxes = number of ways of achieving Case i + number of ways of achieving Case ii = 15 + 10 = 25 ways.
hence the area of all the convex pentagons will be same.
i.e
11111112
Gf who will get 2 gifts can be find out in 8 C 1 ways = 8 ways.
Now 2 gifts can be given to selected gf in 9 C 2 ways. And remaining 7 gifts can be given to remaining 7 gf in 7! ways. So total no of ways= 8 × 9 C 2 × 7! = $$frac{{8 imes left( {9 imes 8}
ight)}}{{2 imes 7!}}$$ = 36 × 8 × 7! = 36 × 8!