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## Clarification needed on combinatorics problem

tagged by: Brent@GMATPrepNow

This topic has 4 expert replies and 2 member replies
knight247 Legendary Member
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#### Clarification needed on combinatorics problem

Fri Nov 21, 2014 10:03 am
In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

OA 16! ÷ (4!)^4

I already got the answer by doing 16C4 * 12C4 * 8C4 * 4C4

My question is, since the four children are NOT identical, shouldn't the above calculation also have a 4C1*3C1*2C1 in there? Considering the four children are NOT identical, we would need to pick one kid each time we need to assign a kid an assortment of four gifts, don't we?

Detailed explanations would be appreciated. Many thanks in advance.

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Fri Nov 21, 2014 10:14 am
knight247 wrote:
In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

I already got the answer by doing 16C4 * 12C4 * 8C4 * 4C4.
Let the four children be Adam, Bobby, Cindy and David.
From 16 gifts, the number of ways to choose 4 to give to Adam = 16C4.
From the remaining 12 gifts, the number of ways to choose 4 to give to Bobby = 12C4.
From the remaining 8 gifts, the number of ways to choose 4 to give to Cindy = 8C4.
From the remaining 4 gifts, the number of ways to choose 4 to give to David = 4C4.
To combine these options, we multiply:
16C4 * 12C4 * 8C4 * 4C4.

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Last edited by GMATGuruNY on Thu Oct 15, 2015 11:11 am; edited 1 time in total

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Brent@GMATPrepNow GMAT Instructor
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Fri Nov 21, 2014 10:17 am
knight247 wrote:
In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?
If we take the task of distributing the 16 gifts and break it into [b]stages, we can see that we need not perform the additional calculations you suggest.

Let's say the children are named A, B, C, and D

Stage 1: Select 4 gifts to give to child A
Since the order in which we select the 4 gifts does not matter, we can use combinations.
We can select 4 gifts from 16 gifts in 16C4 ways (= 16!/(4!)(12!))
So, we can complete stage 1 in 16!/(4!)(12!) ways

Stage 2: select 4 gifts to give to child B
There are now 12 gifts remaining
Since the order in which we select the 4 gifts does not matter, we can use combinations.
We can select 4 gifts from 12 gifts in 12C4 ways (= 12!/(4!)(8!))
So, we can complete stage 2 in 12!/(4!)(8!) ways

Stage 3: select 4 gifts to give to child C
There are now 8 gifts remaining
We can select 4 gifts from 8 gifts in 8C4 ways (= 8!/(4!)(4!))
So, we can complete stage 3 in 8!/(4!)(4!) ways

Stage 4: select 4 gifts to give to child C
There are now 4 gifts remaining
NOTE: There's only 1 way to select 4 gifts from 4 gifts, but if we want the answer to look like the official answer, let's do the following:
We can select 4 gifts from 4 gifts in 4C4 ways (= 4!/4!)
So, we can complete stage 4 in 4!/4! ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 16 gifts) in [16!/(4!)(12!)][12!/(4!)(8!)][8!/(4!)(4!)][4!/4!] ways

A BUNCH of terms cancel out to give us (= 16!/(4!)⁴)

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
- http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html
- http://www.beatthegmat.com/mouse-pellets-t274303.html

MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
- http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
- http://www.beatthegmat.com/combinatorics-problem-t267079.html

DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/ps-counting-t273659.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html

Cheers,
Brent

_________________
Brent Hanneson – Founder of GMATPrepNow.com
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prada Senior | Next Rank: 100 Posts
Joined
08 Dec 2010
Posted:
64 messages
1
Thu Oct 15, 2015 11:01 am
GMATGuruNY wrote:
knight247 wrote:
In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

I already got the answer by doing 16C4 * 12C4 * 8C4 * 4C4.
Let the four children be Adam, Bobby, Cindy and David.
From 16 gifts, the number of ways to choose 4 to give to Adam = 16C4.
From the remaining 12 gifts, the number of ways to choose 4 to give to Bobby = 12C4.
From the remaining 8 gifts, the number of ways to choose 4 to give to Cindy = 8C4.
From the remaining 4 gifts, the number of ways to choose 4 to give to David = 4C4.
To combine these options, we multiply:
16C4 * 12C5 * 8C4 * 4C4.
Hey Mitch on your last line I believe you mean 12C4 and not 12C5?

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GMATGuruNY GMAT Instructor
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Thu Oct 15, 2015 11:14 am
Hey Mitch on your last line I believe you mean 12C4 and not 12C5?
Thanks for pointing out the typo.
I've amended my solution accordingly.

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GMATGuruNY@gmail.com
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Dutta Newbie | Next Rank: 10 Posts
Joined
31 Jan 2016
Posted:
2 messages
Mon Feb 08, 2016 9:54 am
Hey Brent,

Since the 4 kids are not identical should we not consider selecting as to who receives the 1st set of 4 gifts and who the 2nd and so on. So shouldn't the ans be multiplied by a 4!?

Brent@GMATPrepNow wrote:
knight247 wrote:
In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?
If we take the task of distributing the 16 gifts and break it into [b]stages, we can see that we need not perform the additional calculations you suggest.

Let's say the children are named A, B, C, and D

Stage 1: Select 4 gifts to give to child A
Since the order in which we select the 4 gifts does not matter, we can use combinations.
We can select 4 gifts from 16 gifts in 16C4 ways (= 16!/(4!)(12!))
So, we can complete stage 1 in 16!/(4!)(12!) ways

Stage 2: select 4 gifts to give to child B
There are now 12 gifts remaining
Since the order in which we select the 4 gifts does not matter, we can use combinations.
We can select 4 gifts from 12 gifts in 12C4 ways (= 12!/(4!)(8!))
So, we can complete stage 2 in 12!/(4!)(8!) ways

Stage 3: select 4 gifts to give to child C
There are now 8 gifts remaining
We can select 4 gifts from 8 gifts in 8C4 ways (= 8!/(4!)(4!))
So, we can complete stage 3 in 8!/(4!)(4!) ways

Stage 4: select 4 gifts to give to child C
There are now 4 gifts remaining
NOTE: There's only 1 way to select 4 gifts from 4 gifts, but if we want the answer to look like the official answer, let's do the following:
We can select 4 gifts from 4 gifts in 4C4 ways (= 4!/4!)
So, we can complete stage 4 in 4!/4! ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 16 gifts) in [16!/(4!)(12!)][12!/(4!)(8!)][8!/(4!)(4!)][4!/4!] ways

A BUNCH of terms cancel out to give us (= 16!/(4!)⁴)

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
- http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html
- http://www.beatthegmat.com/mouse-pellets-t274303.html

MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
- http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
- http://www.beatthegmat.com/combinatorics-problem-t267079.html

DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/ps-counting-t273659.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html

Cheers,
Brent

### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
Joined
08 Dec 2008
Posted:
11517 messages
Followed by:
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GMAT Score:
770
Mon Feb 08, 2016 1:50 pm
Dutta wrote:
Hey Brent,

Since the 4 kids are not identical should we not consider selecting as to who receives the 1st set of 4 gifts and who the 2nd and so on. So shouldn't the ans be multiplied by a 4!?
We have already accounted for the children being non-identical.
At each stage, we give gifts to a particular child (child A, B, C and D)

Cheers,
Brent

_________________
Brent Hanneson – Founder of GMATPrepNow.com
Use our video course along with

Check out the online reviews of our course
Come see all of our free resources

GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

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