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| Math Central | Quandaries & Queries |
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Question from Jenny, a student: Hi! help please...i'm kinda confused here..i can solve problems like 3 women and 3 men seated alternately but this is definitely giving me a headache... A committee of 15 -- 9 women and 6 men -- is to be seated at a circular table (with 15 seats). In how many ways can the seats be assigned so that no two men are seated next to each other? Is it right to solve this problem in this manner? (6!)(9-1)! = 29,030,400 with the woman on a specific position.. Which one is correct? Or do I still have to add the two results to obtain: 29,030,400 + 43,545,600 = 72,576,000 ??? Please help..thanks! |
Hi Jenny,
Neither solution is correct. It is a good idea to put a particular person in a specific position. The gender of that person should not matter though.
Let's say the person placed at the "head" of teh table is female. That leaves you with 8 females and 6 males. The quantity 8!6! counts the number of ways to arrange the remaining men and women, given that you know the places where they will be sitting. For example, if you know which 8 seats the women will have then there are 8! ways to seat 8 women in them, given that the ninth woman stays in her special seat.
What would happen if you first arranged the nine women in a circle (with one specific woman at the top) and then chose which spaces between consecutive women would have a man in them? Would there be any double counting? Try it with three women and one man, and four women and two men.
Good luck.
Victoria
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