GMATBusters wrote:
GMATbuster's Weekly Quant Quiz#10 Ques #2
For Questions from earlier quizzes: Click HereIn 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!
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*
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