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

Question

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!

nkin · Answer

Books have to be divided equally. So each box should get 12/4 = 3 books only.
So 1st box gets 3 books from 12 in 12C3 ways. Now, 9 books remaining. 
So 2nd box gets 3 books from 9 in 9C3 ways. Now 6 books remaining.
So 3rd box gets 3 books from 6 in 6C3 ways. Now 3 books remaining.
So 4th box gets 3 books from 3 left in 3C3 ways or 1 way.

Ways of distributing 12 books among 4 similar boxes will be 12C3*9C3*6C3*1
Assuming that different boxes means which books go into which box will make a difference, then we will be able to arrange the total ways of distributing 12 books in 4! ways.

So total ways = 12C3*9C3*6C3*1*4!

Option E

prasangrai · Answer

E.
12 different books are there...and we have to divide in 4 boxes. So each group will have 3 books.

So number of ways for selecting 3 books is 12C3*9C3*6C3.

also since the 4 boxes are different we can arrange the group's among them.
So for that number of ways will be four.

So option E

Posted from my mobile device

Shubhranil · Answer

Person 1:
3 books out of 12= 12C3
Person 2:
3 books out of 9= 9C3
Person 3:
3 books out of 6= 6C3
Person 4:
3 books out of 3= 3C3

Thus total possible solutions:
12C3x9C3x6C3x3C3x4!

4! is used because the 4 people can be arranged in 4! ways.

Posted from my mobile device

ankitahyd2013 · Answer

GMATbuster's Weekly Quant Quiz#10 Ques #2 
 For Questions from earlier quizzes:  Click Here

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!

12 different books be distributed equally among 4 different boxed in 12C3*9C3*6C3 ways 
Ways to arrange these books in 4 different boxes is 4!
So the answer is 12C3*9C3*6C3*4!

Sent from my iPhone using  GMAT Club Forum mobile app

rahul6019 · Answer

For distributing 12 books equally across 4 boxes, we need to split 3 books in each.

First 3 books can be selected in 12C3 ways, second 3 is 9C3 ways, third 3 in 6C3 ways and last 3 in 3C3 ways.

Now, let four boxes be A, B, C and D. First box can be selected in 4 ways as it can be any of A, B, C or D. Lets say A is selected as first box. Similarly, Second box can be selected in 3 ways as it can be any from B, C and D and third box can be selected in 2 ways and last box in 1 way.

So, total number of ways = 12C3*9C3*6C3*3C3*4*3*2*1 = 12C3*9C3*6C3*1*4!

Hence, E is correct answer.

BrentGMATPrepNow · Answer

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   =  (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. So, be sure to learn it.

RELATED VIDEOS
 https://www.youtube.com/watch?v=w5d5bvCRyd4

https://www.youtube.com/watch?v=_PFBp2BoMUY

ScottTargetTestPrep · Answer

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

Fdambro294 · Answer

We have 12 distinct items that we want to distribute equally among 4 different boxes.

We can approach the question in stages.

Step #1:
we can divide the different books into identical “stacks.” Assume we have the 12 books on the floor, and we will separate them into 4 groups of 3 books each (before we place them in the box).

We can start out using the combinations formula for each selection.

(12 c 3) (9 c 3) (6 c 3) (3 c 3)

However, there is a slight problem. Because the size of each grouping is identical, we have overcounted the number of “identical stacks.”

(12 c 3) includes all the ways we can take the 12 books and place them into different groups of 3 books.

Let’s label the books:
A - B - C - D - E - F - G - H - I - J -K - L

One of the ways the selections could have ended up is the following:

(A, B, C) —-> chosen when we did (12 c 3)
(D, E, F) —> chosen when we did (9 c 3)
(G, H, I) —> chosen when we did (6 c 3)
(J, K, L)

However, instead of choosing the books (A, B, C) first, we could have chosen the books in this order.

(D, E, F)
(A, B, C)
(G, H, I)
(J, K, L)

Since at this stage, all we are doing is separating the books into identical stacks, we have overcounted. At this stage, we do not want to count the different offerings and shufflings.

For each case such as the above, we have counted 4! ways.

Therefore, to adjust, we divide the calculation by 4! in order to get the number of ways to separate the 12 books into “identical stacks” of 3 books a piece.

(12 c 3) (9 c 3) (6 c 3) (3 c 3)
_______________________
4!

STEP # 2:

Now that we have all the different ways we can make our identical groupings of 3 different books, for each way we need to find out the ways in which we can ARRANGE these stacks into the 4 different boxes.

This part is easy. We can do so in:
4! ways

Notice that when we multiply by *(4!) it Cancels with the (4!) in the denominator and we are left with:

(12 c 3) (9 c 3) (6 c 3) (3 c 3)

Because (3 c 3) = 1

The answer is:

*C*

Posted from my mobile device

I1ya · Answer

You can solve the problem in two ways:

1. Consider 12 books as a 12-character record in which every three letters are the same, aaabbbcccddd, since each book is assigned 1 of 4 boxes (box a, box b, box c, box d). Then the solution is: 12!/3!*3!*3!*3!

2. Consider the number of placement in each of the 4 boxes. For the first box 12C3, for the second 9C3, for the third 6C3, for the fourth 3C3. Then the solution is: 12C3 *9C3 *6C3*3C3

Both solutions are equivalent. Note to the second solution: since any 3 books out of 12 can get into the first box, then any three books can get into the second, third and fourth box. This means that it doesn't matter which box is selected first or last. Multiply by 4! no need.

Answer C.

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards