# Simple Arrangements and Selections - committee question

• Apr 18th 2011, 07:05 PM
DBA
Simple Arrangements and Selections - committee question
How many ways can a committee be formed from 4 men and 6 woman with:

c) 5 people, and not all of the 3 O'Hara sisters can be on the committee.
The answer is C(10,5) - C(7,2)

I do not understand where the C(7,2) comes from. Can someone explain this?

I tried to break it up into committees of 5 people:
0 men - 5 woman ---> I can only choose out of 5 women since one sister of 3 sisters is out
1 men - 4 women --> choose out of 5 women
2 men - 3 women --> choose out of 5 women
3 men - 2 women--> choose out of 6 women, since all 3 sisters cannot be chosen anyway
4 men - 1 women --> choose out of 6 women
5 men - 0 women --> choose out of 6 women

and in the end I add up.
C(4,0)*C(5,5) + C(4,1)*C(5,4) + C(4,2)*C(5,3) + C(4,3)*C(6,2) + C(4,4)*C(6,1)

But I get a different number, so this must be wrong.

Thanks for any help
• Apr 18th 2011, 07:34 PM
DBA
Actually, I can answer it myself. I just figured it out.
The total amout to create a committee with 4 men and 6 women = 10 people is C(10,5)
The forbitten committees have the 3 sisters. So 3 of the 5 places in the committee are occupied by the sisters and we choose only people for the last two places.
We have 7 people to choose from (10 - 3 sisters = 7), so it is C(7,2)

Then we just substract the forbitten committees from the total possible committees C(10,5) - C(7,2)
• Apr 18th 2011, 08:35 PM
Soroban
Hello, DBA!

Seeing your first try, I got curious . . .

Quote:

How many ways can a committee be formed from 4 men and 6 woman with:

c) 5 people, and not all of the 3 O'Hara sisters can be on the committee?

Suppose the three sisters are A, B and C.
And the other three women are x, y and z.

There are five cases to consider . . .

(1) 4 men, 1 woman
. . .Choose all 4 of the men: C(4,4) = 1 way.
. . .Choose any 1 of the 6 women: C(6,1) = 6 ways.
. . .Committees with 4 men and 1 woman: 1 x 6 = 6

(2) 3 men, 2 women
. . Choose 3 of the 4 men: C(4,3) = 4 ways.
. . Choose any 2 of the 6 women: C(6,2) = 15 ways.
. . Committees with 3 men and 2 women: 4 x 15 = 60

(3) 2 men, 3 women
. . Choose 2 of the 4 men: C(4,2) = 6 ways.
. . Choose 3 of the 6 women:
. . . . Choose xyz: C(3,3) = 1 way
. . . . Choose 2 of xyz, 1 of ABC: C(3,2) x C(3,1) = 9 ways
. . . . Choose 1 of xyz, 2 of ABC: C(3,1) x C(3,2) = 9 ways
. . . . There are: 1 + 9 + 9 = 19 ways to choose 3 women.
. . Committees with 2 men and 3 women: 6 x 19 = 114

(4) 1 man, 4 women
. . Choose 1 of the 4 men: C(4,1) = 4 ways.
. . Choose 4 of the 6 women.
. . . . Choose xyz, choose 1 of ABC: C(3,3) x C(3,1) = 3 ways
. . . . Choose 2 of xyz, choose 2 of ABC: C(3,2) x C(3,2) = 9 ways
. . . . There are: 3 + 9 = 12 ways to choose 4 women.
. . Committees with 1 man and 4 women: 4 x 12 = 48

(5) 5 women
. . Choose 3 of xyz, choose 2 of ABC: C(3,3) x C(3,2) = 3 ways
. . Committees with 5 women: 3

Therefore, there are: .6 + 60 + 114 + 48 + 3 .= .231 committees.

. . . ta-DAA!

I've got to get a life . . .