In how many ways can 6 chocolates be distributed among 3 children? A c

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

When distributing 6 identical chocolates among 3 children, we can use "Stars and Bars" method to solve the problem.

Imagine the 6 chocolates as 6 stars in a row. To divide these stars among 3 children, we need 2 bars to create 3 separate sections. For example:

***|*|** would represent that the first child got 3 chocolates, the second got 1, and the third got 2.
||****** would represent that the first child got 0 chocolates, the second got 0, and the third got 10.
****||** would represent that the first child got 4 chocolates, the second got 0, and the third got 2.

So, the problem becomes one of arranging these 6 identical stars and 2 identical bars in a row, which is given by 8!/(6!2!) = 28.

Answer: B.

Hope it helps.

Hi

Multichoosing problem.

We need to distribute 6 identical chocolates among 3 children: (\(x_1, x_2, x_3\) - # each one will get)

\(x_1 + x_2 + x_3 = 6\)

\(_{3+6-1}C_6 = _8C_6 = _8C_2 = \frac{8*7}{2} = 28\)

Answer B

can be solve using stick method,,
imagine 2 sticks and 6 choclates ..
so.. it wud luk like this ||CCCCCC

permutation of the above = (8!/2! 6!) = 28

ans B

i took a bit long route -
3 +ve digit that can sum to 6
006-3 ways
051-6 ways
411-3 ways
321-6 ways
330- 3 ways
222- 1 way
204- 6 ways

total- 28 ways

Let, three children get a, b and c chocolates respectively

i.e. a+b+c = 6

Now, the whole number solutions of an equation = (n+r-1)C(r-1) = (6+3-1)C(3-1) = 8C2 = 28

Answer: Option B

ALTERNATIVE

Let, three children get a, b and c chocolates respectively

i.e. a+b+c = 6

@a=0, (b,c) can be (0,6) (1,5) (2,4) (3,3) (4,2) (5,1) (6,0) i.e. 7 ways
@a=1, (b,c) can be (0,5) (1,4) (2,3) (3,2) (4,1) (5,0) i.e. 6 ways
@a=2, (b,c) can be (0,4) (1,3) (2,2) (3,1) (4,0) i.e. 5 ways
and so on...

Total Ways of distribution = 7+6+5+4+3+2+1 = 28

Answer: Option 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?
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

Hi BrentGMATPrepNow, I don't understand how you are getting 8 letters (IIOOOOOO). We have 6 identical chocolates, so these are the 6 Os given in your solution, but how are you also getting the 2 identical Is? What do these stand for in the initial problem? In other words, how do we have 8 objects when there are only 6 chocolates given?

Thanks, Kaushik

Thanks Bunuel, makes much more sense now! So the way we determine the number of bars is by intuitively thinking through how many groups (in this case people) we need to divide the objects among, and then by observing how the objects can be broken up into such groups? And then, we can also treat the bars as identical objects, so they will be included in the calculation N!/(r1!)*(r2!)...(rn!) where we are arranging N total objects (including the bars) and r1, r2,...rn is the number of times each repeating group repeats? Let me know if I'm understanding this correctly now.

Practice more to understand better.

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