Question

Each of the 3 persons is to be given some identical items such that product of the numbers
of items received by each of the three persons is equal to 30. In How many maximum
different ways can this distribution be done.
a. 21
b. 24
c. 27
d. 33

Answer 1

your answer

the answer is 27 because the the product is 3
which can be written as 3×3×3 =27

Answer 2

Hey Dear,

◆ Answer -

(c) 27

◆ Explanation -

First, We have to find sets of three numbers which whose product is 30 -

(i) A : {1,1,30}

Possible permutations = 3!/2! = 3×2×1 / 2×1 = 3

(ii) B : {1,5,6}

Possible permutations = 3! = 3×2×1 = 6

(iii) C : {1,2,15}

Possible permutations = 3! = 3×2×1 = 6

(iv) D : {2,3,5}

Possible permutations = 3! = 3×2×1 = 6

(v) E : {1,3,10}

Possible permutations = 3! = 3×2×1 = 6

Therefore, no of different ways distribution can be done is -

N = 3 + 6 + 6 + 6 + 6

N = 27 ways

Hope this helped you.