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
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In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 0 to 6
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 GMATinsight
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Distributing 6 identical chocolates among 3 children using permutation and combination to find the whole solution
(n + r  1) C(r  1)
This will give the number of ways n identical chocolates can be distributed for r different children
Given that; n = 6 and r = 3
(n + r  1) C( r  1)
(6 + 3  1) C(3  1)
= 8 C 2
$$now\ u\sin g\ \frac{n!}{r!\left(nr\right)!}$$
$$where\ n\ =\ 8\ and\ r\ =\ 2$$
$$8C2\ =\frac{8!}{2!\left(82\right)!}$$
$$\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$
$$=\frac{8\cdot7}{2}$$
$$=\frac{56}{2}$$
$$8C2=28$$
$$Answer\ =\ B$$
(n + r  1) C(r  1)
This will give the number of ways n identical chocolates can be distributed for r different children
Given that; n = 6 and r = 3
(n + r  1) C( r  1)
(6 + 3  1) C(3  1)
= 8 C 2
$$now\ u\sin g\ \frac{n!}{r!\left(nr\right)!}$$
$$where\ n\ =\ 8\ and\ r\ =\ 2$$
$$8C2\ =\frac{8!}{2!\left(82\right)!}$$
$$\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$
$$=\frac{8\cdot7}{2}$$
$$=\frac{56}{2}$$
$$8C2=28$$
$$Answer\ =\ B$$
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Solution:GMATinsight wrote: ↑Thu Jul 16, 2020 9:02 pmIn 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 obtain the number 6 as the sum of 3 nonnegative integers (where the integers are in nonincreasing order) as follows:
{6, 0, 0}, {5, 1, 0}, {4, 2, 0}, {3, 3, 0}, {4, 1, 1}, {3, 2, 1} and {2, 2, 2}
However, within each set above, we can permute the numbers since, for example, child A gets 6 chocolates, child B gets 0 and child C gets 0 is different from child A gets 0, child B gets 6 and child C gets 0. The numbers in above sets can be permuted in:
3!/2! = 3, 3! = 6, 3! = 6, 3!/2! = 3, 3!/2! = 3, 3! = 6, and 3!/3! = 1 ways, respectively
There are 3 + 6 + 6 + 3 + 3 + 6 + 1 = 28 ways.
Answer: B
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It may seem odd at first, but this question can be reduced to the analogous question: In how many ways can we arrange the letters in IIOOOOOO?GMATinsight wrote: ↑Thu Jul 16, 2020 9:02 pmIn 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
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Let me explain the relationship between this question and the original question.
First, however, let's say the three children are child A, child B, and child C
One possible arrangement of the 8 letters is OOIOIOOO
This arrangement represents child A receiving 2 chocolates, child B receiving 1 chocolate, and child C receiving 3 chocolates
Likewise, the arrangement OIOOOOIO represents child A receiving 1 chocolate, child B receiving 4 chocolates, and child C receiving 1 chocolate
And the arrangement OOIIOOOO represents child A receiving 2 chocolates, child B receiving 0 chocolates, and child C receiving 4 chocolates
Okay, enough examples.
Now that we understand the relationship between the original question and the analogous question, let's see how many ways we can arrange the letters in IIOOOOOO
ASIDE
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

In the case of the letters in IIOOOOOO, . . .
There are 8 letters in total
There are 2 identical I's
There are 6 identical O's
So, the total number of arrangements = 8!/(2!)(6!) = 28
Answer: B
Cheers,
Brent