Problem Solving

sukhman wrote:Gordon buys 5 dolls for his 5 nieces. The gifts include two identical Sun-and-Fun beach dolls, one Elegant Eddie dress-up doll, one G.I. Josie army doll, and one Tulip Troll doll. If the youngest niece doesn't want the G.I. Josie doll, in how many different ways can he give the gifts? [spoiler](5! / 2! - 4! / 2!) = 48.[/spoiler]

Let's first pretend that we DO NOT have two identical Sun-and-Fun beach dolls (we'll deal with that later). So, we'll pretend that there are 5 different dolls to distribute among the nieces.

Take the task of distributing the dolls and break it into stages, beginning with the MOST RESTRICTIVE stage (giving a doll to the youngest niece).

Stage 1: Select a doll for the youngest niece
Since the youngest niece doesn't want the G.I. Josie doll, this stage can be completed in 4 ways

Stage 2: Select a doll for another niece
At this point, there are 4 dolls remaining, so this stage can be completed in 4 ways

Stage 3: Select a doll for another niece
At this point, there are 3 dolls remaining, so this stage can be completed in 3 ways

Stage 4: Select a doll for another niece
This stage can be completed in 2 ways

Stage 5: Select a doll for the last remaining niece
This stage can be completed in 1 ways

By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus distribute all 5 dolls) in (4)(4)(3)(2)(1) ways (= 96 ways)

IMPORTANT: So, if we pretend that there are not two identical Sun-and-Fun beach dolls, then there are 96 possible distributions.
Of course, there ARE two identical Sun-and-Fun beach dolls, so how do we deal with that?

We need to recognize that each unique distribution has been counted twice.
To see what I mean, let's call the two identical dolls SF1 and SF2.
So, in our original solution, one possible distribution might see Sue getting the SF1 doll and Mary getting the SF2 doll, and another distribution might see Sue getting the SF2 doll and Mary getting the SF1 doll (with the other 3 dolls given to the other nieces in the same way). Since the SF1 and SF2 dolls are identical, we are counting the same distribution twice.

To account for this "double counting," we must divide 96 by 2 to get 48

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775