In how many ways 5 different chocolates be distributed to 4 children

Question

In how many ways 5 different chocolates be distributed to 4 children such that any child can get any number of chocolates?

A. 20 
B. 24
C. 120
D. 625
E. 1024

I saw this question on a youtube video and the solution showed 4*4*4*4*4 = 1024, but i was thinking it would be 5*5*5*5 and this answer isn't even in the options. Please help me understand this.

Abhishek009 · Accepted Answer

Let the 5 cholcolates be :  C_1  ,  C_2  ,  C_3  ,  C_4  &  C_5 
And there be 4 Students :  S_1  ,  S_2 ,  S_3  &  S_4

Now, The first Chocolate can be given in 4 ways ( Either to  S_1  or  S_2  or  S_3  or  S_4  )
The Second Chocolate can be given in 4 ways ( Either to  S_1  or  S_2  or  S_3  or  S_4  )
The Third Chocolate can be given in 4 ways ( Either to  S_1  or  S_2  or  S_3  or  S_4  )
The Fourth Chocolate can be given in 4 ways ( Either to  S_1  or  S_2  or  S_3  or  S_4  )
The fifth Chocolate can be given in 4 ways ( Either to  S_1  or  S_2  or  S_3  or  S_4  )

Hence, the total No of ways possible is  4^5  =  1024

Hence, answer will be (E) 1024...

Dennis03 · Answer

Thanks a lot Abhishek!

a70 · Answer

This is a typical example of chocolates going to children. So each chocolate has 4 options and total 5 children. Hence 4^5

turnmlz · Answer

Having the answer in the OG post is not helpful to people using these questions to study.

GmatDaddy · Answer

Lets denote Chocolates by C

So we can distribute C1 in 4 ways
C2-4 ways
C3-4 ways
C4-4 ways
C5-4 ways

Total no of ways of distribution: 4*4*4*4*4
=4^5

An easy way without actually calculating the answer would be to check out he last digit.
Even Powers of 4 yield 4 as unit's digit.
Odd powers of 4 yield 6 as units digit.

Hence Answer: E

ShankSouljaBoi · Answer

https://gmatclub.com/forum/5-rings-on-4 ... er#p649557 check this out

KanishkM · Answer

Another example of players^occurrence

(4)^5

1024

E

CAMANISHPARMAR · Answer

The number of ways in which 5 different chocolates be distributed to 4 children such that any child can get any number of chocolates is 4 * 4 * 4 * 4 * 4 = 1024. Option E is the correct answer

ScottTargetTestPrep · Answer

Since each chocolate can be given to any of the 4 children, each of the 5chocolates has 4 choices. So the total number of ways is 4 x 4 x 4 x 4 x 4 = 4^5 = 1024.

Answer: 5

QuantMadeEasy · Answer

1st chocolate can be given in 4 ways 
2nd can also be given in 4 ways
similarly 3rd, 4th and 5th can be given in 4 ways each
So total ways to give chocolate = 4^5 = 1024

E is correct

Abhi077 · Answer

Each chocolate is distributed in 4 ways and hence if there are 5 children, it will b 4 X 4 X 4 X 4 X 4 = 1024

JHTIPU18 · Answer

Let the 5 cholcolates be : C1C1 , C2C2 , C3C3 , C4C4 & C5C5
And there be 4 Students : S1S1 , S2S2, S3S3 & S4S4

Now, The first Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )
The Second Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )
The Third Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )
The Fourth Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )
The fifth Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )

Hence, the total No of ways possible is 4545 = 1024

Posted from my mobile device

Gustavoncf · Answer

What is wrong with my approach?

We have 5 different chocolates: A,B,C,D,E
And since we need to divide them by 4, and the question states that any child can have any number of chocolates, we have 3 divisions to permute: III

ex: AB | C | D | E

The first kid got AB, the second C, the third D, and the last E

ABCDEIII = 8!/3!

Aren't we assuming that no child can end with zero chocolates with the solutions that were posted?

Bunuel · Answer

Your approach is incorrect precisely because it doesn’t account for the fact that the chocolates are different. For example, the first child receiving A and B is different from the first child receiving D and E, but your method doesn’t consider this distinction.

In the correct approach, 4^5 represents the number of ways each of the 5 different chocolates can be assigned to any of the 4 children. This means that each chocolate has 4 options: it can be "assigned" to child 1, 2, 3, or 4. Therefore, the total number of assignments is 4 * 4 * 4 * 4 * 4 = 4^5. This method accounts for all possible distributions, including cases where one or more children receive no chocolates. For example, if every chocolate is assigned to the first child, all 5 chocolates will go to that child, leaving the others with none.

Hope it helps.

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In how many ways 5 different chocolates be distributed to 4 children

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