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ChrisS
 4 years ago
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?
ChrisS
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(12C2)*(20C3) for the first one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(12C2)*(20C3)(2C2)*(11C1)(19C2) for the second question

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.0To 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
 4 years ago
Best ResponseYou've already chosen the best response.0@Luis, how did you arrive at your answers?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oops the * are suppose to be + , its been a whilte since I did this type of questions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(12C2)+(20C3) (12C2)+(20C3)(2C2)+(11C1)+(19C2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the second should really be (12C2)+(20C3)[(2C2)+(11C1)+(19C2)] sorry, i'm missing brackets
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