Unformatted text preview: Problem 19 (counting committee’s with constraints) 6 , but since two men won’t serve together 3 we need to compute the number of these pairings of three men that have the two that won’t serve together. The number of committees we can form (with these two together) is given by 2 4 · = 4. 2 1 So we have 6 3 − 4 = 16 , 8 3 = 56 diﬀerent groups of women, Part (a): We select three men from six in possible groups of three men. Since we can choose we have in total 16 · 56 = 896 possible committees. Part (b): If two women refuse to serve together, then we will have with these two women in them from the we have 8 3 − 2 2 · 8 3 6 1 2 2 · 6 1 groups ways to draw three women from eight. Thus = 56 − 6 = 50 , 6 3 = 20 ways. In possible groupings of woman. We can select three men from six in total then we have 50 · 20 = 1000 committees. Part (c): We have 8 3 · 1 1 6 3 · total committees, and 7 2 · 1 1 · 5 2 = 210 , committees containing the man and women who refuse to serve together. So we have 8 3 total committees. · 6 3 − 1 1 · 7 2 · 1 1 · 5 2 = 1120 − 210 = 910 , ...
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Penn State.
 Winter '08
 G.JOGESHBABU
 Counting, Probability

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