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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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
$$now\ u\sin g\ \frac{n!}{r!\left(n-r\right)!}$$
$$where\ n\ =\ 8\ and\ r\ =\ 2$$
$$8C2\ =\frac{8!}{2!\left(8-2\right)!}$$
$$\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$
$$=\frac{8\cdot7}{2}$$
$$=\frac{56}{2}$$
$$8C2=28$$
$$Answer\ =\ B$$
(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
$$now\ u\sin g\ \frac{n!}{r!\left(n-r\right)!}$$
$$where\ n\ =\ 8\ and\ r\ =\ 2$$
$$8C2\ =\frac{8!}{2!\left(8-2\right)!}$$
$$\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$
$$=\frac{8\cdot7}{2}$$
$$=\frac{56}{2}$$
$$8C2=28$$
$$Answer\ =\ B$$
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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 non-negative integers (where the integers are in non-increasing 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