Problem Solving

In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

SOURCE: https://www.GMATinsight.com

Answer: option B

GMATinsight wrote:In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

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

To solve the problem above, we need 6 identical chocolates and 2 identical separators, as follows:
OO|OO|OO.
The number of ways to arrange 8 elements composed of 6 identical identical chocolates and 2 identical separators = 8!/(6!2!) = 28.

The correct answer is B.

Since we want to distribute 6 chocolates among 3 children, the exercise is equivalent to find the natural solutions of the equation: x+y+z=6, where x, y and z represent the 3 children.

The solution's sets of this equation are: A={2,2,2}, B={3,2,1}, C={3,3,0}, D={4,1,1}, E={4,2,0} F={5,1,0} and G={6,0,0}.

The sum of the permutations of all the sets above will give us the ways to distribute the chocolates.

We need to know the formula of permutations with repetitions when order is important, which is the case. Let's suppose we have n elements where the first element repeats a times, the second one repeats b times, the third one repeats c times, . . . , i.e. n=a+b+c+. . .

The number of different groups we can make with these n elements is given by the formula: n! / (a!* b!* c!* . . . ).

Now, the number of permutations of the:
- set A is: 3!/3! = 1.
- set B, E and F are: 3!/0! = 6.
- sets C, D and G are: 3!/2!=3.

Thus, there are 1+6+6+6+3+3+3 = 28 ways to distribute the 6 chocolates among the 3 children.

The answer is B.