Question

If you have a group of 6 men and 4 women, how many groups of 5 can you make that have at least one woman?

Answer 1

there will be 2 groups with 5 members each
no. of woman in each group can be
1 and 3
2 and 2
3 and 1

So possible combinations are :

C(4,1)*C(6,4) + C(4,2)*C(6,3) + C(4,3)*C(6,2)

Answer 2

Why are you adding them up to 4? It's a group of 5 members

Answer 3

I did not explain well.
Leme try this way.

in 1 group,either there will be 1 woman and 4 men,
or
2 women, 3 men
or
3 women , 2 men

there can not be 4 women in a group as that will leave other group as 0 women.

Hence possibilities are
C(4,1) * C(6,4)
or
C(4,2) * C(6,3)
or
C(4,3) * C(6,2)

Answer 4

We shall add all them for the final result.

Answer 5

Oh wait, maybe I misunderstood the ques.

Answer 6

There can be 4 women in a group since we are forming just 1 group.hmm
So we will add C(4,4)*C(6,1) as well.

Answer 7

I'll type it as it shows up on the homework I just changed up the values a bit

A committee of 5 people must be chosen from a group of 6 men and 4 women. If the committee is required to have at least 1 woman, how many different committee choices are possible?

Answer 8

one way to go is to compute the chance (x) to have a group without women, and then do 1-x.

Answer 9

But if I'm calculating the number of possible combinations, how would I use chance?

Answer 10

correct, I read the question too fast :-(

Answer 11

Lol it's ok

Answer 12

I think @ataly might have been on the right track... find the total number of possible combinations, and then subtract #groups without women.

Answer 13

Total possible combinations = 10C5

# without women = 6C5

10C5 - 6C5 = 252 - 6 = 246

(not 100% on this, haven't done combinations in a while)

Answer 14

Ok, thanks!

Answer 15

No prob. Hopefully @shubhamsrg or someone will check that. I think it's right though... total groups is:
from 10 people choose 5 people,

groups without women are:
from 6 men choose 5 men.

Answer 16

The long way would be like so:

6C4*4C1 + 6C3*4C2 + 6C2*4C3 + 6C1*4C4 = 246

So both ways get the same answer... which is usually a good sign!