1. ## Money Distribution

How many ways are there to distribute $100 among three people, Bill, Frank and Joe, such that: a) Bill gets at least$10 and Frank gets at least $20? b) Same as above but now we require that Joe gets at most$4?

I am having some trouble getting started on this problem. I would appreciate any suggestions on how to solve this. Thanks

How many ways are there to distribute $100 among three people, Bill, Frank and Joe, such that: a) Bill gets at least$10 and Frank gets at least $20? b) Same as above but now we require that Joe gets at most$4?
You need to post something to show effort on your part.
A) Hint: How many ways can we give $70 to three people? 3. My attempt at a solution for A): suppose A starts off with$10, and B starts off with $20 (so they will already satisfy the 2 conditions) and find how many ways you can distribute the rest of the$70.

A B C
0 0 70
0 1 69
0 2 68
..
..
0 70 0

so when A is fixed with $0, there are 70 ways of distributing the$70 between B and C

A B C
1 0 69
1 1 68
..
..
1 69 0

so when A is fixed with $1, there are 69 ways to distribute the rest of the$69

..
..

continuing the pattern for A fixed with $n, there is 70-n ways to distribute the$(70-n) between B and C

thus, the sum of all these possibilities is 70 + 69 + .. + 2 + 1 = 70(1+70) / 2 = 2485 ways

Does this solution make any sense for A)?
I did something similiar for B)...but am having trouble including the new condition of $4. Thanks for your help 4. The number of ways to put N identical items into k distinct cells is$\displaystyle \binom{N+k-1}{N}=\frac{(N+k-1)!}{N!(k-1)!}\$.
In this problem we have 100 one dollar bills and 3 different people.
Give Bill 10 of them and Frank 20 of them. That leaves 70 of them to deal with.