I have a question.There is a group of 5 men and a group of 7 women.With how many ways can each of the 5 men get married with one of the 7 women?
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Number of ways to choose $5$ out of $7$ woman:
$${7} \choose {5}$$
Number of ways to permute 5 woman to 5 man:
$$ 5!$$
All together:
$${7 \choose{5}}5! =2520 $$
benh
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1And what if it is allowed that a woman can marry simultaneously as many men as she wants and all men have to marry,and the same man cannot be married with two women?? – evinda Apr 24 '14 at 23:33
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Assuming monogamy (No woman is married to more than one man), it is $$7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 = 2520$$ In general, it is for $n$ women and $k$ men $$P(n,k) = \frac{n!}{(n-k)!}$$
AlexR
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1And what if it is allowed that a woman can marry simultaneously as many men as she wants and all men have to marry,and the same man cannot be married with two women?? – evinda Apr 24 '14 at 23:33
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Line the men up in some order (the exact order doesn't make a difference). How many possible choices are there for the first man's wife? The second man's, once the first man's has been chosen? The third man's, once the first two men's have been chosen? Continue in this fashion and multiply the answers at each stage.
Cameron Buie
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1For the first man can marry one of the 7 women,the second 6 women,the third 5 women,the fourth 4 women,and the fifth 3 women.So,it is 76543.Thanks for your answer!!! – evinda Dec 27 '13 at 21:34
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1And what if it is allowed that a woman can marry simultaneously as many men as she wants and all men have to marry,and the same man cannot be married with two women?? – evinda Apr 24 '14 at 23:34
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It would be better to ask that as a separate question. I'm afraid I don't have time to address it. – Cameron Buie Apr 25 '14 at 11:35