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