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

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

Thanks. OA is 70.

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

Oops...

4C1 * 5C2 + 4C2 * 5C1 = 70

Woman Man Man OR Man Woman Woman

When there is OR, we add them as follow:

aj5105 wrote:4C1 * 5C2 + 4C2 * 5C1 = 70

anything wrong with :

9c3 - 4c3 - 5c3 ?

Still gives 70 - but am not sure of the approach

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

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

where did i go wrong ?

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