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## dinner party permutation

A dinner party consisting of 5 couples, sit around a rectangular table, with ladies and gentlemen altering. The host and hostess each occupy one end of the table and their guests are arranged four on each side. Find the number of ways the party can be seated.

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read this: Reflections Vol20, No3, Aug 95

It should give you all the knowledge you need to do these permutations, then go back and do all the problems you posted here for practice, you should be able to.

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With my primitive knowledge I got 4!*4!*2 = 1152

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Wherever the host sits, he has to be surrounded by two female guests, and so the gender arrangement gets completely defined. Because this is a rectangular table, he only has 2 options of where to sit (had this been a circular table, he'd have 10). So wherever he sits, there are only 4 places where female guests can sit and 4 places where male guests can sit. Since order is important, we use factorials:

2*4!*4!=1152 as cesar82 said.

The other possibility is if the couples have to sit together, but the problem didn't specify that explicitly so it seems you have to go with the solution above.

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I agree. In essence this Q is asking "in how many different ways can you seat 4 men" AND "in how many ways can you seat 4 women" = 4! * 4! = 1152

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