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A club has 6 female and 8 male members. A president, VP, secretary and treasurer. In how many ways is this possible if?...

a) an equal number of men and women hold office? b) the president or VP is male?

I have tried counting both equal quantities of men and women (2 for each) by using 6P2 + 8P2. This number is too small. The answer is 10080. I am baffled as how to approach this question.

Jinzu
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    Please add context (where this question comes from; why it's of interest to you) and your related work (what concepts are familiar to you that should apply to this question; anything you might've tried and where it stopped working), etc. – Jonathan Y. Oct 16 '13 at 20:09

3 Answers3

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For the first one, we have to do several things:

  • we must choose two women to hold office
  • we must choose two men to hold office
  • we must choose two positions for the women to hold (which then determines the positions the men may hold)
  • we must assign each woman a position
  • we must assign each man a position

How many ways can each of these things be done? What can we then conclude?

For the second one, I recommend an alternate approach. Figure out the total number of ways the seats can be assigned, then subtract the number of ways that the seats can be assigned where neither president nor VP is male. I leave the former to you. To find the latter, we need to do several things again:

  • we must choose a woman to be president
  • we must choose a woman to be VP
  • we must assign two people to the other positions
Cameron Buie
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  • I do not understand the first answer for a. But I am perplexed as to how you would assign each man and women a position. Also do you multiply the options or add them. – Jinzu Oct 16 '13 at 20:21
  • Well, we have preselected two positions that are to be held by women, for example, and we have chosen two women to fill those positions. In how many ways can we then assign the chosen women to specific positions? – Cameron Buie Oct 16 '13 at 20:38
  • As for multiplication versus addition, think of it this way. Once we've chosen our two women, we can choose the men a certain number of ways, regardless of which women we chose. Once we choose the women and men, we can choose positions for the women to hold in a certain number of ways, regardless of which women and men we chose. So on, until finally we can choose specific positions for the men in a certain number of ways, regardless of preceding choices. Is each choice giving us additional unrelated options, or compounding the options we've already got with more details (like forks in a road)? – Cameron Buie Oct 16 '13 at 20:43
  • Unfortunately I still do not know how to calculate the last three steps for the part a. Thank you for the help. – Jinzu Oct 16 '13 at 22:08
  • There are 4 positions. How many ways can we choose two of them (to be held by women)? Once we've chosen those two (say President and treasurer for example), in how many ways can we assign woman A and woman B to those positions? How many ways can we assign man 1 and man 2 to the other two positions? – Cameron Buie Oct 16 '13 at 23:02
  • Do you understand how to do part b? Did you figure out whether it should be multiplication or addition? – Cameron Buie Oct 16 '13 at 23:03
  • I have not figured out part b, I was working on another subject. I say the way for part a is multiplication. – Jinzu Oct 16 '13 at 23:10
  • It should indeed be multiplication. The idea for part b will be pretty much the same as part a, except we're making different (and fewer) choices. – Cameron Buie Oct 17 '13 at 00:12
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This works as well:

$$\binom{6}{2}\binom{4}{2}\cdot2!\binom{8}{2}\cdot2!=10080$$

First we pick two women, then two jobs, give them order. Then we pick the two men, and give them an order. We don't worry about picking their jobs since there are only two jobs left.

emka
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a) (total) - (woman president AND woman VP) = P(14, 4) - 6*5*12*11 = 10080

so (total positions) - [(the number of women to choose 1st position)(women for 2nd position)(people remaining for third position)(people remaining for 4th position)]

b) (male pres) + (male VP) - (male pres AND male VP) = (8*13*12*11) + (8*13*12*11) - (8*7*12*11) = 20064

male pres = 8 men first choice (times) [13 * 12 * 11 (which represents three people chosen for last three positions)]

male VP same as male pres

male pres AND male VP = 8 men pres * (8-1 men for VP because one taken for pres) * [12*11 (which represents two people chosen for last two choices)]

Jinzu
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