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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


Ahh... I see the issue now, thanks a lot!! And yes indeed the article was from your post... Thanks again!!
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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?

Similar questions to practice:
in-how-many-ways-can-5-different-marbles-be-distributed-in-170689.html
in-how-many-ways-can-5-different-candies-be-distributed-in-141072.html
in-how-many-ways-can-5-different-candiesbe-distributed-among-141070.html
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itsworththepain
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!



we can eliminate right away C and D.
E - 4!*5!=120*24 = 240*12=480*6=960x3 = 2880...way too much.. so out.
B - 5^4 - well, we have 5 fruits and 4 children..each children can be given 1, 2, 3, 4, or 5 fruits. so most likely should be 4^5 than 5^4.
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itsworththepain
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!


Since each child can get 0 to 5 fruits, the number of ways the 5 different fruits be distributed among four children is 4^5.

Answer: A
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I was trying to use the stick method (that looks like this A B | C | D | E) but that didn't work. I suppose it's because the fruits are different not identical?
I could solve the question with the basic 5 slots 4 choices each approach, but I'm trying to understand the stick method and cases where it's applicable. Could someone elaborate on why the stick method didn't work in this case. Does it only work for identical-object cases? Am I missing something obvious here? Thanks
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Hey , thanks all for great post.
I have one query , i am trying to understand the difference between (4 raise to power 5 ) thinking, one fruit can be given to any of 4 children--> 4*4*4*4*4

Also ,(5 raise to power 4) wherein each children can take any of 5 available fruits..

Can anyone please explain the difference..
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CrackVerbalGMAT, why cannot I use the below formula for this question?

(n+r−1) C (r−1)

Is it because objects are not identical?
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KarishmaB
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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


Why don’t we think this way- the first person can choose fruits in 5 ways..second person can aslo choose 5 ways...and so on...then the answer would be (5)^4
I am struggling to understand what's wrong in this thought process.

KarishmaB chetanu
I would be very grateful if you can elaborate this issue.

Posted from my mobile device
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KarishmaB
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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


Why don’t we think this way- the first person can choose fruits in 5 ways..second person can aslo choose 5 ways...and so on...then the answer would be (5)^4
I am struggling to understand what's wrong in this thought process.

KarishmaB chetanu
I would be very grateful if you can elaborate this issue.

Posted from my mobile device


Your approach allows each child to pick a fruit and the same fruit to be picked again by another child.

The question states that each fruit is distributed to a child but doesn't require that each child receive a fruit.
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Asked: 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.)

The number of ways to distribute 5 different fruits among 4 children = 4^5
Since each fruits has 4 choices for distribution.

IMO A
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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.)

Solution : since condition states that some children can recieve 0 and some can recieve more than one fruit. Therefore each child can receive fruits in 6 ways each (0, 1,2,3,4 or 5 =6). Since no fruits to a child is also a possibility. So should we not go with 6×6×6×6 = 6^4 ways?

Posted from my mobile device
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itsworththepain
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?
­The one who doesnt have a choice will- Fruits will go down
as children can get no or all means UP
and the no of times = 5
hence 4^5
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itsworththepain
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?
­the no contraint is on children
so No sits down = 4
and the no of times = 5
hence 4^5
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Why can't we use the partition method here?
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itsworththepain
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?

basic statistics: n children, k fruits.
formula is n^k. So 4^5.
Ans A
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anvigargg
Why can't we use the partition method here?
anvigargg,

Great question! The partition method confusion is one of the most common mistakes in distribution problems. Let me clarify exactly why it doesn't work here.

The Key Distinction: Distinct vs. Identical Objects

The partition method (stars and bars) works ONLY when objects are identical. Here, we have 5 different fruits - they're distinct!
Think about it this way:
  • If we had 5 identical apples → partition method would work
  • But we have (say) apple, banana, orange, mango, grape → each is unique

Why Your Approach Doesn't Work:

When you partition, you're essentially saying "put 2 objects here, 1 object there" etc. But with distinct fruits, which 2 fruits matters! Giving Child A {apple, banana} is different from giving them {orange, mango}.

The Correct Approach:

Since fruits are distinct, think from each fruit's perspective:
  • Apple can go to any of 4 children → 4 choices
  • Banana can go to any of 4 children → 4 choices
  • Orange can go to any of 4 children → 4 choices
  • Mango can go to any of 4 children → 4 choices
  • Grape can go to any of 4 children → 4 choices

Total ways = \(4 \times 4 \times 4 \times 4 \times 4 = 4^5\)

Quick Decision Framework for GMAT:

  1. Objects identical + Distribution → Stars & Bars (partition method)
  2. Objects distinct + Distribution → Multiplication Principle
  3. Objects distinct + Selection → Combinations/Permutations

Remember: On the GMAT, always check if objects are "distinct," "different," "unique" vs. "identical," "same," "indistinguishable" - this determines your entire approach!

Answer: (A) \(4^5\)
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Core Idea (exam brain)

For each fruit, you have 4 choices (which child gets it).

Since fruits are different, each decision is independent.

So total ways:
4^5

Why not others?

5^4 → would mean each child chooses a fruit (wrong structure)

5! → arranging fruits (no ordering here)

4! → arranging children (irrelevant)

4! × 5! → mixing two unrelated permutations
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