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A 3-member rowing team

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A 3-member rowing team

by jsl » Mon Oct 27, 2008 3:07 pm
A 3-member rowing team is to be selected from 4 men and 5 women. How many different 3-member teams be formed subject to the requirement that each team have at least 1 woman and at least 1 man in it?

20
70
80
84
504
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by earth@work » Mon Oct 27, 2008 3:29 pm
no. of ways when 2 women + 3 men in team = 5c2*4c1=40 ways
2men+1 women = 5c1*4c2=30 ways
total=70 ways
ans IMO B=70
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by rohangupta83 » Mon Oct 27, 2008 3:33 pm
total number of ways to select a 3 member team = 9!/3!(9-3)! = 9!/3!6! = 9*8*7/3*2 = 12*7 = 84

number of ways to select a team of 3 men = 4!/3!(4-3)! = 4

number of ways to select a team of 3 women = 5!/3!2! = 5*4/2 = 10

therefore, ways to have a team of at least 1 male and 1 female = 84 - 4 - 10 = 70
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by jsl » Tue Oct 28, 2008 4:24 am
Thanks. OA is 70.

I subtracted 2 instead of subtracting all combinations of M & W.

Oops...
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by aj5105 » Tue Oct 28, 2008 8:57 am
4C1 * 5C2 + 4C2 * 5C1 = 70
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by logitech » Tue Oct 28, 2008 9:00 am
Woman Man Man OR Man Woman Woman

When there is OR, we add them as follow:
aj5105 wrote:4C1 * 5C2 + 4C2 * 5C1 = 70
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by pbanavara » Tue Oct 28, 2008 1:33 pm
anything wrong with :

9c3 - 4c3 - 5c3 ?

Still gives 70 - but am not sure of the approach
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by Stuart@KaplanGMAT » Tue Oct 28, 2008 2:35 pm
pbanavara wrote:anything wrong with :

9c3 - 4c3 - 5c3 ?

Still gives 70 - but am not sure of the approach
There are two general approaches to complex proability/counting questions.

1) add up the things you want; and

2) subtract the things you don't want from the total possibilities.

You chose approach (2), which is perfectly acceptable.

9C3 = total # of 3 person teams
4C3 = # of all male teams
5C3 = # of all female teams

Total # of 3 person mixed gender teams = total # of 3 person teams - # of all male teams - # of all female teams
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by pbanavara » Tue Oct 28, 2008 3:00 pm
Stuart Kovinsky wrote:
pbanavara wrote:anything wrong with :

9c3 - 4c3 - 5c3 ?

Still gives 70 - but am not sure of the approach
There are two general approaches to complex proability/counting questions.

1) add up the things you want; and

2) subtract the things you don't want from the total possibilities.

You chose approach (2), which is perfectly acceptable.

9C3 = total # of 3 person teams
4C3 = # of all male teams
5C3 = # of all female teams

Total # of 3 person mixed gender teams = total # of 3 person teams - # of all male teams - # of all female teams
Thanks Stuart
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by 2009wish » Wed Oct 29, 2008 3:48 am
where did i go wrong ?

5C1 * 4C1 * 7C1 (remaining 3men and 4women - one seat )
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