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

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

[quote="gmatbusters"]
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

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.

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. So, be sure to learn it.

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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

@chethan2u - does the different box mean nothing here in this question ?

Thanks.

Thought that the boxes also must be arranged

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

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.

Bunuel

Shouldn't we multiply the expression 12C3 x 9C3 x 6C3 x 3C3 with 12! coz the books are different

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