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AoPS Counting and Probabilty Help . . . Please?

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My son is trying to do one of the challenge problems, 2.29 in Intro to Counting and Probability. He's stuck, and neither my husband nor I are able to figure out how to unstick him. I'd appreciate any help anyone could offer?


The n members of a committee and numbered 1 through n. One of the members is designated as "the Grand Pooh-bah." The n members sit in a row of n chairs, but no member greater than the Grand Pooh-bah may sit in the seat to the immediate right of the Grand Pooh-bah. Suppose that the Grand Pooh-bah is member number p, where 1 is less than or equal to p is less than or equal to n. Find a formula, in terms of n and p, for the number of ways for the committee to sit.


Using hints from the solutions manual, we got as far as:


(n-1)! + (n-1)(p-1)(n-2)!


But we're supposed to simplify, and I'm completely confused. I know what the solutions manual says the answer should be, but we can't figure out how to get there.


I know there are many smart and mathy folks here. So, help please?

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Yay for fun math!


I followed the solution up to your result, and I agree so far. Here's what I'd do to simplify it.


(n-1)! + (n-1) (p-1) (n-2)! =


First note that (n-1)! = (n-1) (n-2) (n-3) ... 1 = (n-1) (n-2)!, so


= (n-1)! + (p-1) (n-1)!


= [(n-1)!] [1 + (p-1)] by factoring out the common term


= (n-1)! [ p] by simplifying the last term


So I get p(n-1)! as the final answer.

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