In how many ways can 16 different gits be divided among four

[AAPL]

Magoosh

In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

A. \(16^4\)

B. \((4!)^4\)

C. \(\frac{16!}{(4!)^4}\)

D. \(\frac{16!}{4!}\)

E. \(4^{16}\)

OA C

[Brent@GMATPrepNow]

Magoosh

In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

A. \(16^4\)

B. \((4!)^4\)

C. \(\frac{16!}{(4!)^4}\)

D. \(\frac{16!}{4!}\)

E. \(4^{16}\)

OA C

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 D
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!)�

Answer: C

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

[Ian Stewart]

If you know that in any counting situation, when the order of k things doesn't matter, you can first pretend order does matter and then divide by k!, then you can just: imagine putting all 16 gifts in a row, which you can do in 16! ways. Give the first four gifts to the oldest child, the next four gifts to the next oldest, and so on. Now for each of the four kids, the order of their four gifts doesn't matter, so we need to divide that 16! by 4! four times, and the answer is 16! / (4!)^4.

[Scott@TargetTestPrep]

Magoosh

In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

A. \(16^4\)

B. \((4!)^4\)

C. \(\frac{16!}{(4!)^4}\)

D. \(\frac{16!}{4!}\)

E. \(4^{16}\)

OA C

The first child can choose any 4 gifts from the 16 gifts; thus, (s)he has 16C4 ways to choose them. Once (s)he has chosen the 4 gifts, the second child can choose any 4 gifts from the remaining 12 gifts; thus (s)he has 12C4 ways to choose them. Likewise, the third child has 8C4 ways to choose his or her 4 gifts, and the last child has 4C4 ways to choose his or her 4 gifts. Thus the total number of ways the 16 gifts can be divided among the four children such that each child will receive 4 gifts is:

16C4 x 12C4 x 8C4 x 4C4

(16 x 15 x 14 x 13)/4! x (12 x 11 x 10 x 9)/4! x (8 x 7 x 6 x 5)/4! x (4 x 3 x 2 x 1)/4!

(16 x 15 x 14 x 13 x ... x 4 x 3 x 2 x 1)/(4! x 4! x 4! x 4!)

16!/(4!)^4

Answer: C