sagarsameer wrote:
Question: In how many ways can 5 different fruits be distributed among four children? (Some children may get more than one fruit and some may get no fruits.)
(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!
Correct Answer is 1024 and I understand the reasoning behind that.
But the approach I followed was:
Say we keep the 5th fruit aside and distribute the other 4 fruits among the 4 children
No of ways to distribute 4 fruits among 4 children = 4!
Now for each of these 4! Combinations, 5th fruit can be distributed to any of the 4 children
i.e. 4 new combinations for each of the 4! combinations
No of ways to distribute 5th fruit = 4*4!
5th fruit can be selected in 5 different ways
Total combinations are 5*4*4! = 480
What’s wrong here? What is it that I am missing here? What are the other 1024-480 combinations that I am missing?
I believe you picked up this question from my blog post so you know how to correctly solve it. So I will focus on only pointing out the error you committed. To start off, how do you get "No of ways to distribute 4 fruits among 4 children = 4! "? You are not given that each child gets only one fruit. If each child were to receive only one fruit, then you could have said that the first fruit can be given away in 4 ways, 2nd fruit in 3 ways and so on to get 4*3*2*1.
Some children may get none so others may get 2 or 3 or 4. Each fruit can be given out in 4 ways so 4*4*4*4*4 = 1024
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Karishma
Veritas Prep GMAT Instructor
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56% (01:15) correct 
