In how many ways can 6 chocolates

[BTGmoderatorDC]

In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 1 to 6 and all the chocolates are identical.

A) 10
B) 15
C) 21
D) 28
E) 56

Can some experts show the best solution in this problem?

OA A

[GMATGuruNY]

In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 1 to 6 and all the chocolates are identical.

A) 10
B) 15
C) 21
D) 28
E) 56

We can apply the SEPARATOR METHOD, which I describe here:
https://www.beatthegmat.com/combinations-t120668.html

To guarantee that each child receives at least 1 chocolate, first give 1 chocolate to each child.
At this point, 3 of the 6 chocolates must still be distributed.
Thus, we need 3 identical chocolates and 2 identical separators, as follows:
OOO||.
The number of ways to arrange 5 elements composed of 3 identical identical chocolates and 2 identical separators = 5!/(3!2!) = 10.

The correct answer is A.

Alternate approach:

Let the 3 children be A, B and C.
To guarantee that each child receives at least 1 chocolate, first give 1 chocolate to each child.
Now WRITE OUT the different ways the remaining 3 chocolates can be distributed among the 3 children :
A =1 chocolate, B = 1 chocolate, C = 1 chocolate
A = 2 chocolates, B = 1 chocolate
A = 2 chocolates, C = 1 chocolate
B = 2 chocolates, A = 1 chocolate
B = 2 chocolates, C = 1 chocolate
C = 2 chocolates, A = 1 chocolate
C = 2 chocolates, B = 1 chocolate
A = 3 chocolates
B = 3 chocolates
C = 3 chocolates
Total ways = 10.

The correct answer is A.

[[email protected]]

Hi lheiannie07,

We're told that 6 IDENTICAL chocolates will be distributed among 3 children with each child receiving 1-6 chocolates. We're asked for the total number of ways to distribute the chocolates. Since there aren't that many chocolates, you might find it easiest to 'brute force' this question by just mapping out the various possibilities.

To start, each child has to receive at least 1 chocolate, so the number of possible 'combinations of chocolate' to the 3 children are limited:

1 piece/1 piece/4 pieces - there are 3 ways for this to occur (depending on which of the 3 children receives 4 pieces).
1 piece/2 pieces/ 3 pieces - there are 6 ways for this to occur (again, depending on who gets 3 and who gets 2).
2 pieces/2 pieces/2pieces - there is 1 way for this to occur

3+6+1 = 10 total ways to distribute the identical chocolates.

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 1 to 6 and all the chocolates are identical.

A) 10
B) 15
C) 21
D) 28
E) 56

Let the children be A, B and C. So A can get 1, B can get 1 and C can get 4 chocolates. Of course, this is different from A gets 4, B 1 and C 1, or, A gets 1, B 4 and C 1.

In the calculations below, we will show how 3 positive integers can sum to 6 and the number of ways the 3 numbers can be rearranged among A, B and C (for example, the first calculation below describes the distribution of the 6 chocolates mentioned above):

1 + 1 + 4 = 6

3!/2! = 3 ways

1 + 2 + 3 = 6

3! = 6 ways

2 + 2 + 2 = 6

3!/3! = 1 way

Therefore, there are a total of 3 + 6 + 1 = 10 ways 6 chocolates can be distributed to 3 children.

Answer: A