AAPL wrote:Princeton Review

Coach Miller is filling out the starting lineup for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions: 1 goalkeeper, 2 on defense, 2 in midfield, and 1 forward. Only 2 of the boys can play goalkeeper, and they cannot play any other positions. The other boys can each play any of the other positions. How many different groupings are possible?

A. 60
B. 210
C. 2580
D. 3360
E. 151200

OA D

Since only 2 boys can play goalkeeper, the number of ways to choose a goalkeeper is 2C1 = 2. Since all the other boys can each play any of the other positions, the number of ways to choose 2 defense players is 8C2 = (8 x 7)/2 = 28. After the 2 defense players are chosen, the number of ways to choose 2 midfield players is 6C2 = (6 x 5)/2 = 15. After that, the number of ways to choose 1 forward is 4C1 = 4. Therefore, the total number of ways to choose 6 starters is:

2 x 28 x 15 x 4 = 56 x 60 = 3360

Answer: D

AAPL wrote:Princeton Review

Coach Miller is filling out the starting lineup for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions: 1 goalkeeper, 2 on defense, 2 in midfield, and 1 forward. Only 2 of the boys can play goalkeeper, and they cannot play any other positions. The other boys can each play any of the other positions. How many different groupings are possible?

A. 60
B. 210
C. 2580
D. 3360
E. 151200

OA D

Divide the complete task into 4 stages:
1. Select a goalkeeper. We must select one boy from 2. We can accomplish this in 2 ways.
2. Select 2 for defence. We must select 2 boys from the remaining 8. We can accomplish this in 8C2 ways (28 ways.)
(Note: I'm assuming that order doesn't matter here. That is, there is no left defence and right defence; they are simply on defence)
3. Select 2 for midfield. We must select 2 boys from the remaining 6 boys. We can accomplish this in 6C2 ways (15 ways.)
4. Select 1 for forward. We must select 1 boy from the remaining 4 boys. We can accomplish this in (4 ways.)
The total number of ways to complete the entire task is 2x28x15x4 = 3360 (D)