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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?

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  1. anonymous
    • 4 years ago
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    (12C2)*(20C3) for the first one

  2. anonymous
    • 4 years ago
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    (12C2)*(20C3)-(2C2)*(11C1)(19C2) for the second question

  3. ChrisS
    • 4 years ago
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    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

  4. ChrisS
    • 4 years ago
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    @Luis, how did you arrive at your answers?

  5. anonymous
    • 4 years ago
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    oops the * are suppose to be + , its been a whilte since I did this type of questions

  6. anonymous
    • 4 years ago
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    (12C2)+(20C3) (12C2)+(20C3)-(2C2)+(11C1)+(19C2)

  7. ChrisS
    • 4 years ago
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    Thank you!

  8. anonymous
    • 4 years ago
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    the second should really be (12C2)+(20C3)-[(2C2)+(11C1)+(19C2)] sorry, i'm missing brackets

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