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

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AAPL: Moderator; Posts: 2647; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Mon Mar 04, 2019 5:32 am

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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

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Source: Beat The GMAT — Problem Solving | Mon Mar 04, 2019 5:32 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Mar 04, 2019 6:17 am

AAPL wrote: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!)â�´

Answer: C

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Mon Mar 04, 2019 12:35 pm

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.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8154; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Mar 06, 2019 6:50 pm

AAPL wrote: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

Scott Woodbury-Stewart
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