BTGmoderatorDC wrote: ↑Wed Jun 01, 2022 4:18 am
In how many ways can one divide 12 different chocolate bars into four packs of 3 bars each?
A) 12!/3!4!
B) 12!/(4!)^2
C) 12!/4!(3!)^4
D) 12!/4!(4!)^3
E) 12!/4!(3!4!)
OA
C
Source: Veritas Prep
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!)⁴
Answer: C
Brent Hanneson - Creator of GMATPrepNow.com
