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In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 May 2017, 06:49
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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 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 May 2017, 07:50
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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 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

SOURCE: http://www.GMATinsight.com


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vitaliyGMAT
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 May 2017, 07:06
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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 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

SOURCE: http://www.GMATinsight.com


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
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 May 2017, 09:35
2
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
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 May 2017, 09:46
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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 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

SOURCE: http://www.GMATinsight.com


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
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  09 Jan 2018, 08:15
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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 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

SOURCE: http://www.GMATinsight.com


hi

since chocolates are identical, and there are 3 children, we can think this way

c | c | c c c c

thus

8!
______
6! 2!

= 28, the answer
thanks

cheers, and do consider some kudos, guys
:cool:
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saabdulkhader
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  10 Jan 2018, 00:32
Using stick method,

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

So B
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  23 Jul 2019, 09:29
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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 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

SOURCE: http://www.GMATinsight.com


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


To attend the !!!FREE!!! Live class on Permutation and combination on 24th July 2019 @9 PM IST, click and register on the LINK

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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  25 Jul 2019, 06:13
Can someone explain the stick method, please?
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  28 Jul 2019, 07:22
vitaliyGMAT wrote:
GMATinsight wrote:
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

SOURCE: http://www.GMATinsight.com


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

Why 3+6-1? Can someone explain please?
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Luca1111111111111
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  01 Aug 2019, 04:08
GMATinsight wrote:
GMATinsight wrote:
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

SOURCE: http://www.GMATinsight.com


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


To attend the !!!FREE!!! Live class on Permutation and combination on 24th July 2019 @9 PM IST, click and register on the LINK

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Hello, can you please explain why the whole number solutions of the equation = (n+r-1)C(r-1)?
Thank you
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 Jul 2020, 07:55
Hi Guys,

If chocolates were different, I think that we could use the following Method

3*3*3*3*3*3 =3^6

Is there a way to move from this scenario to actual solution?

Could you please explain the rationale behind a different approach ?

Posted from my mobile device
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 Jul 2020, 07:58
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Albymana wrote:
Hi Guys,

If chocolates were different, I think that we could use the following Method

3*3*3*3*3*3 =3^6

Is there a way to move from this scenario to actual solution?

Could you please explain the rationale behind a different approach ?

Posted from my mobile device


This can not be corrected because this method includes some cases once, some cases TWICE and some cases thrice and the repetition in this case can not be eliminated

I hope this help!
Albymana
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 Jul 2020, 08:13
GMATinsight wrote:
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

SOURCE: http://www.GMATinsight.com


Asked: 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.

C C C C C C ||

Number of ways = 8C2 = 28

IMO B
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Albymana
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  16 Jul 2020, 09:52
GMATinsight wrote:
Albymana wrote:
Hi Guys,

If chocolates were different, I think that we could use the following Method

3*3*3*3*3*3 =3^6

Is there a way to move from this scenario to actual solution?

Could you please explain the rationale behind a different approach ?

Posted from my mobile device


This can not be corrected because this method includes some cases once, some cases TWICE and some cases thrice and the repetition in this case can not be eliminated

I hope this help!
Albymana


Got your point thank you. The rationale behind is that

- when object are different you have to consider the single elements in the allocation (I.e. You consider two different ways when Child 1 receive C1, C2, C5 and child 2 receive C2, C1 and C5

- when object are identical, instead, you only have to care about # of element allocated to students

Am I right?

Best
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BrentGMATPrepNow
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Re: In how many ways can 6 chocolates be distributed among 3 children? A c [#permalink] New post  03 Dec 2020, 17:35
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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 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

SOURCE: http://www.GMATinsight.com


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
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