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The part where the number of seats are limited is causing a difficulty. Seating all of the men and women with this restriction is easy. But the restriction on number of seats mixed it all up.

androlinux
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The condition that no two women sit together warrants that you pick m number of men out of 10 men and n women out of 10 women such that n = 1, 2 or 3 or 4.

For n = 1: The number of ways you can choose 1 of 10 women is ${10\choose1}$ and the rest of the six men are chosen in ${10\choose6}$. Having chosen this, now we need to make sure that these women don't sit together. For n = 1, it is straight 7 ways you can arrange the woman and 6! ways you can arrange the men in remainder of the six seats. Thus the total arrangement is ${10\choose1}{10\choose6}7.6!$

For n = 2: In a similar argument.The number of ways you can choose 2 of 10 women is ${10\choose2}$ and the rest of the five men are chosen in ${10\choose5}$. Having chosen this, now we need to make sure that these women don't sit together.we want to fill $2$ women around $5$ men, then we have a total number of arrangements that is ${(5+1)\choose 2}. 2! 5!$. Thus the total arrangement is ${10\choose2}{10\choose5}{6\choose2}2!.5!$

For n= 3:In a similar argument.The number of ways you can choose 3 of 10 women is ${10\choose3}$ and the rest of the 4 men are chosen in ${10\choose4}$. Having chosen this, now we need to make sure that these women don't sit together.we want to fill $3$ women around $4$ men, then we have a total number of arrangements that is ${(4+1)\choose 3}. 3! 4!$. Thus the total arrangement is ${10\choose3}{10\choose4}{5\choose3}3!.4!$

For n = 4: In a similar argument.The number of ways you can choose 4 of 10 women is ${10\choose4}$ and the rest of the 3 men are chosen in ${10\choose3}$.we want to fill $4$ women around $3$ men, then we have a total number of arrangements that is ${(3+1)\choose 4}. 4! 3!$. Thus the total arrangement is ${10\choose4}{10\choose3}{4\choose4}4!.3!$

The final answer is sum up all cases.

Satish Ramanathan
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  • Be careful. There can be four women if you seat them in the first, third, fifth, and seventh seats. – N. F. Taussig Nov 08 '17 at 10:03
  • You are right that is the fourth case. Will edit – Satish Ramanathan Nov 08 '17 at 10:05
  • What would happen if the men and women are replaced with something that can repeat in the arrangement, say "task" and "break" respectively, other conditions being the same (no two breaks can be together but tasks and breaks can be repeated)? – androlinux Nov 08 '17 at 12:03
  • The first two terms and the last two terms would not exist in each case. In other words, if you have 10 red (men) and 10 blue(women) (idential). you can imagine there is no selection that are identified in the first two terms and last two terms would vanish as there are no more permutations. – Satish Ramanathan Nov 08 '17 at 12:20
  • They are not identical, they are distinct but can still be repeated. https://math.stackexchange.com/questions/2510569/how-many-ways-are-there-to-arrange-10-distinct-tasks-of-type-a-and-10-distinct-t – androlinux Nov 08 '17 at 12:49
  • Could you put in terms of something that I understand. I don't get it – Satish Ramanathan Nov 08 '17 at 12:52
  • I've given a link to the question in my previous comment as the question on this page is already answered. – androlinux Nov 08 '17 at 12:54
  • In the seven seats, how could you repeat something if they are distinct – Satish Ramanathan Nov 08 '17 at 12:54
  • Let me rephrase, Suppose there are 2 set of tasks A and B, and for example one task from set A is "drink water". Then I can drink water multiple times. – androlinux Nov 08 '17 at 12:57