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
SOURCE: https://www.GMATinsight.com
Answer: Option A
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In how many ways can 6 chocolates be distributed among 3 chi
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GMATinsight
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GMATinsight wrote: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
SOURCE: https://www.GMATinsight.com
Answer: Option A
Each child must get at least one chocolate, that means the maximum a child can receive is 4.
3 ways a child can get 4 chocolates.
Each of these is paired with the two other children getting one, so 3 ways.
3 ways a child can get 3 chocolates. One remaining child gets 1 choco, the other gets 2. Two children, therefore 2 ways to achieve this. Total: 3x2 = 6 ways
Each child can receive 2 choco, so 1 way.
No need to go further, since the remaining approaches already covered.
Total number of ways: 3 + 6 + 1= 10
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GMATGuruNY
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We can apply the SEPARATOR METHOD, which I discuss here:GMATinsight wrote: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
https://www.beatthegmat.com/inserting-st ... 67423.html
To guarantee that no child is left empty-handed, give each child a chocolate.
Now use the separator method to count the number of ways to distribute the remaining 3 chocolates.
To solve, we need 3 identical chocolates and 2 identical separators, as follows:
O|O|O.
The number of ways to arrange 5 elements composed of 3 identical chocolates and 2 identical separators = 5!/(3!2!) = 10.
The correct answer is A.
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Scott@TargetTestPrep
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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.GMATinsight wrote: ↑Tue May 16, 2017 7:03 amIn 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
SOURCE: https://www.GMATinsight.com
Answer: Option A
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 that 6 chocolates can be distributed to 3 children.
Answer: A
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