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In how many ways can a committee of five be selected from seven women and four men: 1) if there are no restriction? 2) if there must be exactly one woman on the committee?

I read several times question but still can't understand 1) answer is 462 and 2) answer is 7

thanks in advance!

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    For general comments on counting, see [this previous question and answer](http://math.stackexchange.com/questions/11307/how-do-i-do-combinatorics-counting/11312#11312)2011-10-02

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  1. There are 4+7=11 total people; you're choosing 5 out of that pool of candidates. How many ways are there to choose five out of eleven people? First imagine that the order you choose them in matters (this is not the case, but we'll analyze it first):
    • For the first choice, you have 11 possible people to pick; for the second choice, you have 10 possible people to pick from, that number being independent of the first choice; for the third choice, you have 9 possible to pick from, that number independent of the first two choices, etc. So the total number of ways of picking 5 people out of 11 total is 11*10*9*8*7, or 11!/6!.
    • Now in the above argument, every possible set of five people (e.g. persons A,B,C,D,E) is counted multiple times, indeed, every possible ordering of them is counted. The number of ways to order 5 objects is 5*4*3*2*1 (same as our previous reasoning), so every 5-person set of choices we counted 5! times when we really need to only count them once each, so we divide our total count by that in order to get: 11!/(5!*6!)=462.
  2. Split the selection into two independent parts (do you see how they're independent?):
    • choosing exactly 1 woman out of 7 female candidates, and
    • choosing 5-1=4 men out 4 male candidates.
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    1) 2 ways to choose- because there is only 11, so if you choose 2 times it is 10. Do you mean like this? thx2011-10-02
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    @Sb Sangpi: What do you mean choose 2 times? You need need to choose *five* people out of *eleven* total. Has your class gone over the concept of "[n choose k](http://en.wikipedia.org/wiki/Binomial_coefficient)" yet? (If not, I'll need to make my answer more elaborate.)2011-10-02
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    Yea.I don't know about "n choose k". Can you please elaborate more? I appreciate it. Sorry for my lack knowledge.2011-10-02
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    +1 great answer. thx! Looking forward help! God bless you!2011-10-02
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    2) I can't get the answer **7**2011-10-03
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    @Sb Sangpi: For the first part, there are seven women, and you must pick one. Every choice you could make corresponds to exactly one of the women, and there are seven women, so there must be seven choices. So the number of ways to pick one out of seven women is seven. The other four committee members must be men, and there are exactly four men to choose from. There is only one way to pick four men out of four men: you must pick all of them. So the total number of ways of picking 1 out of 7 female and 4 out of 4 male candidates is 7x1=7.2011-10-03
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    ok! so, let say, if only one men on the committee? there are four man, so there must be 4 choices.so the number of ways to pick one out of 4 men is 4.The other seven committee members must be women, and there are 3 way to pick. so the total number of ways of picking 1 out 4 men and 4 out of 7 women candidate is $7*3=21$ ? but the answer is **140**2011-10-03
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    @Sb Sangpi: Please don't ask about different questions *without telling me what the question is*. However, I did figure it out: you're to choose 1 out of 4 men and 3 out of 7 women. The problem is you're confusing the number of women you choose with the number of ways to choose them: these are not the same! So: how many ways can you choose 3 out of 7 women? My answer shows you how to calculate the number of ways to choose 5 out of 11 people; apply that same type of reasoning in this situation. Once you have the number of ways to choose 3 out of 7 women, multiply that by 4 (from choosing men).2011-10-03