dblok79 wrote:A dog sledder will form a team by choosing six dogs from among the eight he owns. The dog-sledding team consists of three rows of two dogs each. A variation in which different dogs are selected for each row is considered a different team. How many different teams can be formed?
(A)28
(B)56
(C)252
(D)280
(E)2,520
Let's assume that "Fido on the left and Rover on the right"
is different from "Fido on the right and Rover on the left"
We can solve this question using the Fundamental Counting Principle (FCP)
First, let's assign a number to each of the 6 places on the team:
1 2
3 4
5 6
Now, we'll take the task of building a team and break it into stages.
Stage 1: Select a dog for position #1
There are 8 dogs to choose from, so this stage can be accomplished in
8 different ways.
Stage 2: Select a dog for position #2
There are now 7 dogs to choose from, so this stage can be accomplished in
7 ways.
Stage 3: Select a dog for position #3
There are now 6 dogs to choose from, so this stage can be accomplished in
6 ways.
Stage 4: Select a dog for position #4
There are now 5 dogs to choose from, so this stage can be accomplished in
5 ways.
Stage 5: Select a dog for position #5
There are now 4 dogs to choose from, so this stage can be accomplished in
4 ways.
Stage 6: Select a dog for position #6
There are now 3 dogs to choose from, so this stage can be accomplished in
3 ways.
By the Fundamental Counting Principle (FCP) we can complete all 6 stages (and thus fill all 6 positions) in
(8)(7)(6)(5)(4)(3) ways ([spoiler]= 20160 ways[/spoiler])
Okay, since 20,160 is not one of the answer choices. It must be the case that "Fido on the left and Rover on the right"
is the same as "Fido on the right and Rover on the left." My solution for that assumption is next.
Cheers,
Brent
Aside: For more information about the FCP, we have a free video on the subject:
https://www.gmatprepnow.com/module/gmat-counting?id=775
