Permutation And Combination - Study Mode

[#56] 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?
Correct Answer

(B) 3

Explanation

Solution: Let the number of Rose plants be a .
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.

[#57] How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?
Correct Answer

(B) 12 C 4 × 4 C 3 × 7!

Explanation

Solution: 4 consonants out of 12 can be selected in, 12 C 4 ways.
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!

[#58] In how many ways can a committee of 4 people be chosen out of 8 people?
Correct Answer

(B) 70

Explanation

Solution: Required number of ways $$eqalign{
& = {}^8{C_4} cr
& = frac{{8 imes 7 imes 6 imes 5}}{{4 imes 3 imes 2 imes 1}} cr
& = 70 cr} $$

[#59] In how many different ways can the letters of the word EXTRA be arranged so that the vowels are never together?
Correct Answer

(C) 72

Explanation

Solution: Taking the vowels (EA) as one letter, the given word has the letters XTR (EA), i.e., 4 letters. These letters can be arranged in 4! = 24 ways The letters EA may be arranged amongst themselves in 2 ways. Number of arrangements having vowels together = (24 × 2) = 48 ways Total arrangements of all letters = 5! = (5 × 4 × 3 × 2 × 1) = 120 Number of arrangements not having vowels together = (120 - 48) = 72

[#60] In how many different ways can the letters of the word ‘BAKERY’ be arranged?
Correct Answer

(C) 720

Explanation

Solution: The letters of the word 'BAKERY' be arranged in 6! ways = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720