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

Is there anyway that we could move the meeting to a different time? I work from 2:30-5:30 monday through thursday... You think maybe we could do it sometime in the evening on some day instead?

I can do whenever is easy. I work nights, but can move my schedule around some if needed. If there are more then just you and I, then it's whenever it is easiest for us all. Maybe monday nights?

Monday nights can work ...I work til 7pm Tues Wed and Thurs, so if we did meet then I wouldn't be able to make it.

I've got a review session tonight for an exam in a different math class, but let me know when working on this would be the best. Tomorrow after 7 is acceptable with me. I won't be at the Union unless someone else wants to meet there. Last week I was there for 2 hours alone and don't want to waste time preparing myself and walking over there unless it's needed. My cell is 317-605-6720 and I can be reached there at any point in time. Just let me know if you (or anyone reading this) wants to work on the current homework. I never do amazingly and would love the help.


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

Section 5.5

10. For this problem we can assume that order doesn't matter right?

Any suggestions for #10f ? I know what to do for all the croissants except for the 'no more than 3 broccoli croissants.

when it says no more than 3, you can look at it as the previous conditions and picking 3 broccoli croissants, then the previous conditions and picking 2 broccoli croissants etc...

You can also look at it like this: Choose the necessary croissants. Next find the number of ways to choose no more than 3 broccoli for the remaining croissants by subtracting the number of ways to choose more than 3 broccoli for the remaining croissants from the number of ways to choose the remaining croissants with no conditions.

18. How do we account for the overcount?

The overcount comes from the individual decimal digits are indistinguishable. Think of what happens in a string when you switch a 1 and another 1. The string does not change.

This is similar to counting the number of ways of rewriting the word "MISSISSIPPI", which we looked at in class. Of course, in this case, it's not letters that repeat themselves, but digits; the rewritings form "clusters" of size equal to the number of times each digit appears.


I am looking to check my answer on this one. I got 78, did anyone else get something different?

yes, I got something different. remember that it is less than or equal to 11 not just equal to 11. look at the hint. it should help a lot.

I'd also like to check my answer... I got 364. It seems like what I did was almost too easy; if I'm right, then they practically told us how to solve it right in the hint (which is fine by me, lol). Did others get this answer, too?

Yes I got this answer, you can brute force to prove this. x1+x2+x3 = (0...11), and you sum them all up will give you 364 as well.



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.

I don't even understand answer on the book.

This one, if you solved a) part correctly. Is just another 12! multiply by whatever you have in part a. Before that you didn't care about the order of the book as they are chosen. Now each time you pick a book, you take it away from your available choices.

For part A, think of the books as 12 stars, and the shelves as 3 bars dividing the 12 books into the 4 shelves. The rest of this calculation goes back to our work with stars and bars. For part B, use the hint to your benefit. When you are placing the first book, b1, how many positions are there to put it in? 4: to the right of 1, to the right of 2, to the right of 3, to the right of 4. Now, for b2, how many positions are there? 5: now you can additionally choose between to the right of b1 or to the left of b1, whereas before there was just one choice there. Continue this thought process--you should see a pattern develop. Hope that helps. -BD


Does anyone have an idea or a hint to help me on this one? It seems like we talked about this in class but for some reason i can not find it in my notes.

as following the hint, you treat chosen books as bars, so that the question can be converted in the standard model we talked about during class, that is how many ways are there to fill 6 spaces around 5 bars with 7 books, so that there is at least 1 book in the second, third , fourth and fifth spaces. the question can then be modelized in to how many nonnegative integer solutions the equation X1+X2+......+X6=7 has such that X2,X3,X4,X5 are greater or equal to 1


is distributing five distinguishable objects into three indistinguishable boxes the same as 5 indistinguishable objects into three distinguishable boxes? No there, is a difference because the distinguish vs. indistinguish is a 3 vs 5 instead of 5 vs. 3. Did anyone else get 21 as an answer for this one. It is kind of confusing for me especally with the way they describe it in the book.

I got 41 for my answer. I just followed the formula they used in the book. I though it was confusing too so i tested the example in the book with the formula to make sure i was doing it right. ---RPK


Anyone know how to do this one?

look at example 11 in section 5.5

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.

I tested this out for (x+y+z)^2 and used m=3 and n=2. When you multiply it out by hand you get 6 different terms and C(3+2-1,2-1)= C(4,2)= 6. So I don't think you need to worry about the combined terms. I think it accounts for it. The equation you have above x1+x2+x3+...+xm=n can also be thought of as the number of different combinations the exponents can be that all add up to n. Like with the binomial thereom, the term is always there just it may not show because the exponent could be 0 resulting in the term to equal 1. Thinking about it this way makes sense that there would be no need to think of overcount with the combined terms.

Does anyone know when he is going to start graph theroy. And how are his midetrms??

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett