How many different groupings possible?

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Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

by Vemuri » Sun Mar 22, 2009 5:24 am

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 defence, 2 in midfield, and 1 forwaard. 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. 2,580
D. 3,360
E. 151,200

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Source: Beat The GMAT — Problem Solving | Sun Mar 22, 2009 5:24 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

Re: How many different groupings possible?

by Brent@GMATPrepNow » Sun Mar 22, 2009 7:24 am

Vemuri wrote: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 defence, 2 in midfield, and 1 forwaard. 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. 2,580
D. 3,360
E. 151,200

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)

Brent Hanneson - Creator of GMATPrepNow.com

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Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: How many different groupings possible?

by Vemuri » Mon Mar 23, 2009 12:56 am

Brent Hanneson wrote: 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)

Spot on !!!

Brent, how would you solve the problem if order matters in the defence & forward?

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Mar 23, 2009 7:57 am

Brent, how would you solve the problem if order matters in the defence & forward?

Since order would then matter, those stages would require permutations:
Stage 2 would be 8P2 ways (56)
Stage 3 would be 6P2 ways (30)

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

by Vemuri » Mon Mar 23, 2009 9:42 am

Thanks Brent.

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vishaljainsxc: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Sun Jan 30, 2011 6:34 amFollowed by: 1 member

by vishaljainsxc » Wed Sep 28, 2011 11:57 pm

why is the answer not 2C1 x 8C5..since out of the 8 ppl we have to select 3 people

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studentps2011: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Fri Jul 15, 2011 5:37 am; Thanked: 3 times

by studentps2011 » Thu Sep 29, 2011 9:40 am

vishaljainsxc wrote:why is the answer not 2C1 x 8C5..since out of the 8 ppl we have to select 3 people

We can select 5 people from 8 (8C5) to get 56 groups. But then, the 5 boys in each group can be arranged in (5!/(2!x2!) = 30)ways.

So total number of groupings = 2x56x30 = 3360

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