Clarification needed on combinatorics problem

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.

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.

Your solution is correct.
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.

prada wrote: Hey Mitch on your last line I believe you mean 12C4 and not 12C5?

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 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 ([spoiler]= 16!/(4!)�[/spoiler])

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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: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent