Bunuel wrote:

Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?

A. 24

B. 70

C. 210

D. 420

E. 1,680

There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.

1) Row 1: A-B, Row 2: C-D

2) Row 1: A-C, Row 2: B-D

3) Row 1: A-D, Row 2: B-C

4) Row 1: B-C, Row 2: A-D

5) Row 1: B-D, Row 2: A-C

6) Row 1: C-D, Row 2: A-B

Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.

Answer: D

Why have we done 4,5 and 6 selection?What i can discern from the question is different pairings form different teams. So if AB is paired once whether in row 1 or 2, its pairing is done.

Please explain what is wrong in my interpretation of the question?