In how many ways can 6 chocolates be distributed among 3 chi

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Tue May 16, 2017 7:03 am

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

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Answer: Option A

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Tue May 16, 2017 9:43 am

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: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue May 16, 2017 10:17 am

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

We can apply the SEPARATOR METHOD, which I discuss here:
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: GMAT Instructor; Posts: 8210; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: In how many ways can 6 chocolates be distributed among 3 chi

by Scott@TargetTestPrep » Sat Apr 11, 2020 8:09 am

GMATinsight wrote: ↑
Tue May 16, 2017 7:03 am
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

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 that 6 chocolates can be distributed to 3 children.

Answer: A

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