In how many ways can you distribute 10 euros among 5 people by using 1, 2, 5, 10, 20, 50 cents and 1 and 2 euros (with an unlimited supply of every coin).
In part A of the question it doesnt matter what kind of coins you give the people.
At first I thought I could use the possible combinations of getting $a+b+c+d+e=10000$.
But this can only be used when $a , ... , e$ can have every value which is not the case.
Then I thought about using generating functions: Since every person gets 2 euros the number of ways for every person to get the 2 euros should be the same I assume we are looking for the coefficient of $[x^{200}]$, and the generating function is:
$$g(x)=(x^0+x^1+x^2+...)(x^0+x^2+x^4+...)(x^0+x^5+...)(x^0+x^{10}+...)(x^0+x^{20}+...)(x^0+x^{50}+...)(x^0+x^{100}+...)(x^0+x^{200}+...)$$
To find the coefficient here we can rewrite this as: $$g(x) = \frac{1}{1-x}\frac{1}{1-x^{2}}\frac{1}{1-x^{5}}\frac{1}{1-x^{10}}\frac{1}{1-x^{20}}\frac{1}{1-x^{50}}\frac{1}{1-x^{100}}\frac{1}{1-x^{200}}$$
For which we can make a combinatorial expression
$$g(x)=\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{k}\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{2k}\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{5k}\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{10k}\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{20k}\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{50k}\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{100k}\sum_{k=0}^{\infty}\binom{n+k-1}{k}x^{200k} $$
But to find the coefficient of $x^{200}$ here wil take ages. Which makes me rethink my strategy. There must be some possible simplification somewhere. Does anyone have an idea here?? Maybe a variation on the first appoach?
For question B the choice of the coins does matter per person (for instance person A doesnt want any 1 cent coins). So for this problem I will have to use the generating function.. But still how..?
Any help will be appreciated!
I wrote a little program. Make an array ways with indices from $0$ to $1000$ and fill it with $1$s, representing that there is one way to give somebody any amount from $0$ to $1000$ cents using $1$ cent coins. Then for each index $2$ and above in order let ways[i]=ways[i]+ways[i-2] which says you can either give them $i$ cents with $1$ and $2$ cent coins by giving them $i-2$ cents and adding a $2$ cent coin or by not using a $2$ cent coin. Then starting at $5$ let ways[i]=ways[i]+ways[i-5]. Keep going for each type of coin. You could do this in $1000$ lines and nine columns of a spreadsheet with copy down. I get $321335886$ both ways.