A committee of five people is to be chosen from a club that boasts a membership
of 10 men and 12 women. How many ways can the committee be formed if it is
to contain at least two women? How many ways if, in addition, one particular
man and one particular woman who are members of the club refuse to serve
together on the committee?

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- ChrisS

- schrodinger

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- anonymous

(12C2)*(20C3) for the first one

- anonymous

(12C2)*(20C3)-(2C2)*(11C1)(19C2) for the second question

- ChrisS

To get the total combinations without restriction of any kind, I would use (22C5) right?
If that is the case I don't see the first answer because that gives me a number larger than the total without restriction.
(22C5)=22,334
(12C2)*(20C3)=75240

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## More answers

- ChrisS

@Luis, how did you arrive at your answers?

- anonymous

oops the * are suppose to be + , its been a whilte since I did this type of questions

- anonymous

(12C2)+(20C3)
(12C2)+(20C3)-(2C2)+(11C1)+(19C2)

- ChrisS

Thank you!

- anonymous

the second should really be (12C2)+(20C3)-[(2C2)+(11C1)+(19C2)]
sorry, i'm missing brackets

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