# 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

• Generating functions would be the concept you'd apply in general here as Adriano's answer uses this concept. – JB King Aug 7 '13 at 21:30
• Are the fish distinguishable? That is, does it matter that I got fish #2, #7, and #11, or just that I got three fish? – MJD Aug 7 '13 at 21:33
• The fish are not distinguishable. I see I have got two very good answers, thank you all! – Tore Aug 7 '13 at 21:37

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: $$f(x)=\underbrace{(1+x^2+x^4+\ldots)}_{AB}\underbrace{(x^2+x^3+x^4+\ldots)}_{C}\underbrace{(1+x+x^2+\ldots)}_{D}$$

• \begin{align} f(x)&=\frac1{1-x^2}\frac{x^2}{1-x}\frac1{1-x}\\ &=\frac{x^2}{1-2x+2x^3-x^4}\\ &=x^2+2x^3+4x^4+6x^5+9x^6+12x^7+16x^8+20x^9+25x^{10}+30x^{11}+\dots \end{align} – robjohn Aug 7 '13 at 22:09
• Thanks for the answer! I understand everything you have done up to your last sentence. But how did you go from = (x^2)/(1-2x+2x^3-x^4) to =x^2+2x^3+4x^4+...? Im sorry if it´s a stupid question :) – Tore Aug 8 '13 at 8:17
• @Tore, split into partial fractions and use $(1 - z)^{-r} = \sum_{n \ge 0} (-1)^n \binom{-r}{n} z^n = \sum_{n \ge 0} \binom{n + r - 1}{r - 1} z^n$. – vonbrand Aug 26 '15 at 1:14

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 $10+8+6+4+2=30$.