In how many ways can one divide twelve different chocolate

AAPL wrote:Veritas Prep

In how many ways can one divide twelve different chocolate bars equally among four boys?

A. \(\frac{12!}{3!}\)
B. \(\frac{12!}{4!}\)
C. \(\frac{12!}{(3!)^4}\)
D. \(\frac{12!}{(4!)^3}\)
E. \(\frac{12!}{3!4!}\)

OA C

Let the 4 boys be: A, B, C and D

Choose 3 chocolate bars for boy A. Since the order in which we can select the 3 bars doesn't matter, we can use combination.
We can select 3 bars from 12 bars in 12C3 ways.
12C3 = (12)(11)(10)/(3)(2)(1)

Choose 3 chocolate bars for boy B. There are now 9 bars remaining.
We can select 3 bars from 9 bars in 9C3 ways.
9C3 = (9)(8)(7)/(3)(2)(1)

Choose 3 chocolate bars for boy C. There are now 6 bars remaining.
We can select 3 bars from 6 bars in 6C3 ways.
9C3 = (6)(5)(4)/(3)(2)(1)

Choose 3 chocolate bars for boy D. There are now 3 bars remaining.
We can select 3 bars from 3 bars in 3C3 ways.
9C3 = (3)(2)(1)/(3)(2)(1)

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 12 bars) in [(12)(11)(10)/(3)(2)(1)][(9)(8)(7)/(3)(2)(1)][(6)(5)(4)/(3)(2)(1)][(3)(2)(1)/(3)(2)(1)]

Simplify product to get: 12!/(3!)�

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
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- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
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- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
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- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

AAPL wrote:Veritas Prep

In how many ways can one divide twelve different chocolate bars equally among four boys?

A. \(\frac{12!}{3!}\)
B. \(\frac{12!}{4!}\)
C. \(\frac{12!}{(3!)^4}\)
D. \(\frac{12!}{(4!)^3}\)
E. \(\frac{12!}{3!4!}\)

OA C

We see that each boy gets 12/4 = 3 chocolate bars. Therefore, there are 12C3 ways to choose 3 chocolate bars for the first boy, 9C3 ways for the second boy, 6C3 ways for the third boy and 3C3 ways for the last boy. So the total number of ways to divide the 12 chocolate bars equally among 4 boys is:

12C3 x 9C3 x 6C3 x 3C3

= (12 x 11 x 10)/3! x (9 x 8 x 7)/3! x (6 x 5 x 4)/3! x (3 x 2 x 1)/3!

= (12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)/(3! x 3! x 3! x 3!)

= 12!/[(3!)^4]

Answer: C