dblok79 wrote: A variation in which different dogs are selected for each row is considered a different team.

Hmmm, this sentence is confusing.
Are we saying that, within a certain row, the order of the 2 dogs matters or doesn't matter?
That is, are we saying that "Fido on the left and Rover on the right" is different from (or the same as) "Fido on the right and Rover on the left"???

Cheers,
Brent

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

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

Since my last solution didn't match any of the answer choices, let's assume that "Fido on the left and Rover on the right" is the same as "Fido on the right and Rover on the left"

We can solve this question using the Fundamental Counting Principle (FCP)

We'll take the task of building a team and break it into 3 stages.

Stage 1: Select 2 dogs for the first row
Since the order of the selected dogs does not matter, this is a combination.
There are 8 dogs and we must choose 2.
So this stage can be accomplished in 8C2 = 28 different ways.

Aside: If anyone is interested, we have a free video on calculating combinations (like 8C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Select 2 dogs for the second row
There are 6 dogs remaining and we must choose 2.
So this stage can be accomplished in 6C2 = 15 different ways.

Stage 3: Select 2 dogs for the lsat row
There are 4 dogs remaining and we must choose 2.
So this stage can be accomplished in 4C2 = 6 different ways.

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus create a team) in (28)(15)(6) ways ([spoiler]= 2520 ways[/spoiler])

Answer = E

Cheers,
Brent

dblok79 wrote:Awesome Thank you Brent! I have a question:

How do I discern when to use the above strategy vs. stop after stage 1 of the proposed solution?

I went the following route. How many ways can I select a team of 6 dogs from the group of 8?

8C6 = 28. i.e 28 different team configurations of 6. I ended there since I felt each configuration would be different and therefore have a row of 2 that would be different from the next configuration.

If I had looked at it as 8C2 instead of 8C6 I may have gone down the road proposed. How can I avoid this error next time? any ideas?

Thanks

You were still on the right track.
Always ask, "Can I take the task of building possible outcomes and break it into individual stages?"

Your approach begins with...
Stage 1: Select the 6 dogs for the team.
Great, this can be accomplished in 8C6 ways = 28 ways.

Now we need to place each dog in row 1, 2 or 3.

Stage 2: Select 2 dogs for row #1.
There are 6 dogs, so this can be accomplished in 6C2 ways = 15 ways.

Stage 3: Select 2 dogs for row #2.
There are 4 dogs remaining, so this can be accomplished in 4C2 ways = 6 ways.

Stage 4: Select 2 dogs for row #3.
There are only 2 dogs remaining, so there's only one way to accomplish this stage. (By the way, 2C2 = 1)

By the Fundamental Counting Principle (FCP) we can complete all 4 stages in (28)(15)(6)(1) ways (= 2520 ways)

Answer = E

Cheers,
Brent