In how many ways can 12 different books be distributed equally among 4 different boxes?

BTGModeratorVI wrote: ↑

Wed Apr 22, 2020 11:06 am

In how many ways can 12 different books be distributed equally among 4 different boxes?

A) 12C3
B) 12C4
C) 12C3*9C3*6C3
D) 12C4*8C4
E) 12C3*9C3*6C3*4!

Answer: C
Source: E-gmat

So, the 12 distinct books are to be equally distributed among 4 different boxes.

So each box will have 12/4 = 3 books

Three books can be distributed among 4 boxes in the following way...

• The 1st box will get 3 books from 12 books in 12C3 ways.
• The 2nd box will get 3 books from the remaining 9 books in 9C3 ways.
• The 3rd box will get 3 books from the remaining 6 books in 6C3 ways.
• The 4th box will get 3 books from the remaining 3 books in 3C3 ways i.e in only 1 way.

Total no. of ways = 12C3*9C3*6C3*1 = 12C3*9C3*6C3

The correct answer: C

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑

Wed Apr 22, 2020 11:06 am

In how many ways can 12 different books be distributed equally among 4 different boxes?

A) 12C3
B) 12C4
C) 12C3*9C3*6C3
D) 12C4*8C4
E) 12C3*9C3*6C3*4!

Answer: C
Source: E-gmat

Take the task of distributing the books and break it into stages.

We must place 3 books in each of the 4 boxes. So, let's call the boxes box #1, box #2, box #3 and box #4

Stage 1: Select 3 books to go in box #1
Since the order in which we select the books does not matter, we can use combinations.
We can select 3 books from 12 books in 12C3 ways
So, we can complete stage 1 in 12C3 ways

Stage 2: Select 3 books to go in box #2
There are 9 boxes remaining.
So, we can complete this stage in 9C3 ways

Stage 3: Select 3 books to go in box #3
There are 6 boxes remaining.
So, we can complete this stage in 6C3 ways

Stage 4: Select 3 books to go in box #4
There are 3 boxes remaining.
So, we can complete this stage in 3C3 ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 12 books) in (12C3)(9C3)(6C3)(3C3) ways

Check the answer choices....our answer doesn't seem to be there.
However, if we recognize that 3C3 = 1, we can see that [color=blue(12C3)(9C3)(6C3)(3C3)[/color] = (12C3)(9C3)(6C3)(1) = (12C3)(9C3)(6C3)

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 this free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

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


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/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

BTGModeratorVI wrote: ↑

Wed Apr 22, 2020 11:06 am

In how many ways can 12 different books be distributed equally among 4 different boxes?

A) 12C3
B) 12C4
C) 12C3*9C3*6C3
D) 12C4*8C4
E) 12C3*9C3*6C3*4!

Answer: C
Source: E-gmat

Each box will contain 3 books. There are 12C3 ways to put 3 books into the first box, 9C3 ways to put 3 books into the second box, 6C3 ways to put 3 books into the third box, and 3C3 ways to put the last 3 books into the fourth box. Therefore, the total number of ways is:

12C3 x 9C3 x 6C3 x 3C3 = 12C3 x 9C3 x 6C3

Answer: C