A quadruple date
4 pairs of husband and wife went out for a group date one night. How many different seating can be arranged for them if they already booked a row of 8 seats in the cinema but refused to sit next to someone of the opposite gender if they are not their spouses?
The answer is 816.
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1 solution
Thank you, Mark. I was just annoyed to find this same question having another answer, and that was the only reason I reask. Too lazy to type and report those anymore. Thanks again.
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Which question? I will report it.
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https://brilliant.org/problems/2002-math-osp-number-17/ I solved it purely out of curiosity, which of the seating patterns he counted.
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Wow, that's quite exhaustive case checking.
Since it's an Olympiad problem, I believe that there should be a "nicer" way of solving it. I think the principle of inclusion and exclusion (pie) could work well here.
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Well, it was fairly easy to count the through options of " x women on the left", subdivided (if necessary) by "then y men next". There is one option each with x = 4 and x = 3 , there are three options with x = 2 (for y = 4 , 3 , 2 ) and four options with x = 1 (one each for y = 4 , 3 and two for y = 2 ). Counting the number of arrangements for each of these patterns is straightforward enough.
Splitting a combinatorics problem into a small number of subcases which are easy is standard enough Olympiad technique. Perhaps the IEP could be used to subdivide by the number of husband/wife pairs sitting next to each other, but I am not sure it would be any quicker/easier.
I did it much the same way as Mark in regards to the 9 M/F patterns, but the number of possibilities for each pattern is more simply thought of (in my opinion) in the following way:
If we position, say, the four women first, there are 4! ways of doing so. Any M next to a female is then already determined because he must be next to his wife. This leaves a certain number (possibly zero) of men to arrange. If this number of men is m then there are m! ways of arranging them, giving 4!m! arrangements. This leads to:
4!3! 4!2! 4!2!
4!1! 4!1! 4!2!
4!1! 4!0! 4!1!
Well... It SEEMED simpler in my head, but maybe not so simple to explain! (-:
No woman can be sitting between two men, and no man can be sitting between two women. Thus there are 9 seating patterns of the sexes which have a woman on the far left. These are W W W W M M M M W W M M M W W M W M M M W W W M W W W M M M M W W W M M W W M M W M M W W M M W W W M M M M W W W M M M M W W W W M M W W W M M In each case the underlined pairs must be married couples. These 9 cases can be obtained in the following number of ways: 4 × 3 ! × 3 ! 4 × 3 × 2 4 × 3 × 2 4 × 3 × 2 × 2 4 × 3 × 2 4 × 3 × 2 4 × 3 × 2 × 2 4 × 3 × 2 × 2 4 × 3 × 2 each, making a total of 4 0 8 ways so far. There are an equal number of seating arrangements with a man at the far left. Thus the total number of seating arrangements is 8 1 6 .