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Four male and five female members of a commission are to sit in a meeting on a round table. How many ways they can sit in the meeting if:

  1. All male members are to sit together?
  2. All female are to sit together and there is one seat for the chair of the commission. Assume each member is qualified to chair the commission?
  3. If no two male members sit together.
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closed as off-topic by uniquesolution, Rohan, MathOverview, Claude Leibovici, GNUSupporter 8964民主女神 地下教會 Jan 18 '18 at 13:32

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  • 2
    $\begingroup$ Are you in the same class as this guy? $\endgroup$ – Arthur Jan 18 '18 at 11:39
  • $\begingroup$ Nope............ $\endgroup$ – user522659 Jan 18 '18 at 11:45
  • $\begingroup$ @user522659 does my answer help? $\endgroup$ – QuIcKmAtHs Jan 18 '18 at 12:09
  • $\begingroup$ @xcoderx yup man tysm for your help $\endgroup$ – user522659 Jan 18 '18 at 12:10
  • $\begingroup$ Please mark the tick @user522659 and upvote $\endgroup$ – QuIcKmAtHs Jan 18 '18 at 12:11
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  1. If all males were to sit together, that would mean the women were seating together too, since it is a round table. Hence, the number of ways would be the number of ways to arrange 4 men multiplied by the number of ways to arrange 5 women, which is $4!*5!=2880$.

  2. We group the 5 females together, as a block, so now there are 5 blocks, so we have 5! ways of arranging these 5 blocks. Now among the men, there are 4! ways since there are 4 men. The probability of a person selected as chairperson is $^9C_1=9$. So we have $9*4!*5!=25920$ ways.

  3. If no 2 male members sit together, we first consider the case of 4 women and 4 men. Then, the number of ways to arrange the men is 4!, and the number of ways to arrange the women is 4! too. Do note that the men and women here alternate seats. Then our answer would be $4!*4!$. However, now we have 5 women. We need to multiply $^5C_4=5$ to choose 4 women. Then the last women can be paired with any other women, which has 4 ways. So we have $4!*4!*4*5=11520$ ways.

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