Permutation And Combination - Study Mode

[#51] 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?
Correct Answer

(D) 2 × 18!

Explanation

Solution: n objects can be arranged around a circle in (n - 1)!
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!

[#52] 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.
Correct Answer

(B) 63 C 4

Explanation

Solution: Let assume that there are 64 identical balls which are to be arranged in 5 different compartments (Since a, b, c, d, e are distinguishable) If the balls are arranged in a row
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.

[#53] 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?
Correct Answer

(C) 4 × 3 4

Explanation

Solution: The remainder on the first card can be 0, 1, 2 or 3 i.e. 4 possibilities. The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier). For each value on the card the remainder can have 3 possible values. The total number of possible sequences is: 4 × 3 4

[#54] 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?
Correct Answer

(D) 91

Explanation

Solution: We first count the number of committee in which (i). Mr. Y is a member (ii). the ones in which he is not case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
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.

[#55] 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?
Correct Answer

(B) 360

Explanation

Solution: Two horses A and B, in a race of 6 horses . . . A has to finish before B. If A finishes 1 . . . . . B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways = 5 × 4! If A finishes 2 . . . . . B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways = 4 × 4! If A finishes 3 rd . . . . . B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways = 3 × 4! If A finishes 4 th . . . . . B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways = 2 × 4! If A finishes 5 th . . . . B has to be 6th and the top 4 positions could be filled in 4! ways. A cannot finish 6 th , since he has to be ahead of B. Therefore total number of ways: = 5 × 4! + 4 × 4! + 3 × 4! + 2 × 4! + 4! = 120 + 96 + 72 + 48 + 24 = 360