Difficult Combinatorics I have a question I have been trying to solve for some time now:
4 people (let´s call them A, B, C and D) are out fishing. They catch 11 fish. How many ways can the fish be distributed if A and B has to have the same amount of fish, and C has to get at least 2 fish?
Is it possible to make a formula for this problem?
Thanks
Tore
 A: Pretend that $A$ and $B$ are Siamese twins, and they really just count as one person (call this person $AB$). Then this mega person must have an even amount of fish.
One way to solve this problem is to use generating functions by finding the coefficient of $x^{11}$ in the function:
$$
f(x)=\underbrace{(1+x^2+x^4+\ldots)}_{AB}\underbrace{(x^2+x^3+x^4+\ldots)}_{C}\underbrace{(1+x+x^2+\ldots)}_{D}
$$
A: A and B get the same number of fish, so let's call that number $x$.
C has to get at least 2 fish, so let's call the number of fish C gets $y+2$.
Finally, D gets z fish, and there's no constraints on $z$.
So now, since the total number of fish is 11, we have $2x+(y+2)+z=11$, or $2x+y+z=9$. We're interested in the solutions where all three of $x$, $y$, and $z$ are nonnegative integers. Solving for $z$, $z=9-2x-y$, so any solution with $x$, $y$ nonnegative such that $2x+y \le 9$ will give a valid choice. 
We can now just list these. For $x=0$, $y$ can be from $0$ to $9$, giving $10$ possibilities. For $x=1$, $y$ can be from $0$ to $7$. Similarly, for $x=2$, $y$ can go from $0$ to $5$, and for $x=3$, $y$ can go from $0$ to $3$. For $x=4$ the only possible values of $y$ are $0$ or $1$. So the number of ways is $10+8+6+4+2=30$.
