Revision as of 08:42, 10 February 2010 by Jvizza (Talk | contribs)

I would like to meet on Tuesdays from 3-5 and work on this homework as a group. As of right now, meeting in the Union is the best meeting point. I'll be seated next to Starbucks and will bring my book. Please call or text me at 317-605-6720 with any questions or comments. Hopefully we can work through some of these problems together. Ryan Hossler

HW4MA375S10

5.5 - 10, 18, 20, 24, 44, 46, 50, 54, 62

Section 5.5

10.



18. How do we account for the overcount?



20.



24.



44.

Anyone have any ideas for 44b? This could be way off, but is the answer 4*12!^12 because there are 12! ways to place the books in specific order on each shelf, 12 places on the shelves for books (there can be any number on each shelf, but only 12 books can be used), and 4 shelves for the books to rest on?

Look at question 45, then the answer to it in the back of the book.


46.



50.



54.



62. does anyone have hints on this? I think I might have gotten part-way through this. The total number of terms equals the number of solutions to x1+x2+...+xm = n, because in each term, the exponents have to add up to n. Then you could do the stars and bars method, where you have m-1 bars and n stars. That leads you to C(n+m-1, m-1). However, this is where I am stuck; I have no idea how to account for the terms that can be combined. I was thinking maybe none can be combined? But probably not because the question mentioned them.


I can't help much because I too am stuck but I have gotten to the same point you have so I believe it just takes a little tinkering with the data to get the final answer.


Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood