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From: markp@tekmdp.UUCP (Mark Paulin)
Newsgroups: net.math
Subject: Re: Homework problem
Message-ID: <2219@tekmdp.UUCP>
Date: Wed, 14-Sep-83 12:44:30 EDT
Article-I.D.: tekmdp.2219
Posted: Wed Sep 14 12:44:30 1983
Date-Received: Thu, 15-Sep-83 12:16:03 EDT
Lines: 25


In how many ways may 2r people be chosen from n married couples in such a way
that we have exactly k married couples among the 2r people?  k <= r <= n.


1)  Let C(a, b) denote the binomial coefficient "a above b".  Then first choose
    k couples from the n couples -- this may be done in C(k, n) ways.

2)  Next choose, from the remaining (n - k) couples, the 2(r - k) couples which
    will be represented among the 2r people to be assembled.  This may be done
    in C(2(r - k), (n - k)) ways.

3)  Finally, for each of the 2(r - k) couples chosen in 2), choose the member to
    be included among the 2r people.  This may be done in 2 ways per couple, or
    a total of 2**(2(r - k)) ways.

Thus the total number of ways to form the required group of 2r people is:

C(n, k) * C(2(r - k), (n - k)) * 2**(2(r - k))

///.


Mark Paulin
...tektronix!tekmdp!markp