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
In how many ways can 6 chocolates be distributed among 3
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We can apply the SEPARATOR METHOD, which I describe here:BTGmoderatorDC 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
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.
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Let the children be A, B and C. So A can get 0, B can get 0 and C can get 6 chocolates. Of course, this is different from A gets 6, B 0 and C 0, or, A gets 0, B 6 and C 0.BTGmoderatorDC 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
In the calculations below, we will show how 3 nonnegative 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):
0 + 0 + 6 = 6 ---> 3!/2! = 3 ways
0 + 1 + 5 = 6 ---> 3! = 6 ways
0 + 2 + 4 = 6 ---> 3! = 6 ways
0 + 3 + 3 = 6 ---> 3!/2! = 3 ways
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 x 3 + 6 x 3 + 1 = 9 + 18 + 1 = 28 ways 6 chocolates can be distributed to 3 children.
Answer: B
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