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?

Question

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?

anonymous · Answer

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

anonymous · Answer

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

chriss · Answer

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

chriss · Answer

@Luis, how did you arrive at your answers?

anonymous · Answer

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

anonymous · Answer

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

chriss · Answer

Thank you!

anonymous · Answer

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