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I came across this questions:

Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish. How many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the 5 types?

I know how the answer is $\binom{10+5-1}{5-1}=4$ This is just star and bar (type 2 on wikipedia) problem with $n=10$ and $r=5$. But I am not able to guess how I can solve the same with generating functions?

I felt it will be coefficient of $x^10$ in $(1+x+x^2+...)^5$. I calculated it as follows: $(1+x+x^2+...)^5=\sum_{j=0}^\infty\binom{10+j-1}{10-1}x^j$
Coefficient of $x^{10}$ in this is $\binom{10+10-1}{10-1}=\binom{19}{9}=92378$ which definitely does not look correct.

When I checked wolfram alpha it was saying coefficient of $x^10$ in $(1+x+x^2+...)^5$ is $0$.

Whats going on here?

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    $\begingroup$ ${14 \choose 4} = 1001$, not $4$. $\endgroup$ – Travis Willse Sep 26 at 5:45
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The correct generating function here is $$(1+x+ \cdots + x^{10})^5 = \left(\frac{1-x^{11}}{1-x} \right)^5$$

Now, use

  • $\frac{1}{(1-x)^5} = \sum_{n=0}^{\infty}\binom{n+4}{4}x^n$

You get $$[x^{10}]\left(\frac{1-x^{11}}{1-x} \right)^5 = [x^{10}]\sum_{n=0}^{\infty}\binom{n+4}{4}x^n = \binom{10+4}{4} = 1001$$

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  • $\begingroup$ Yess got my silly mistake. Here if we consider $(1+x+x^2+...)^5$ instead of $(1+x+x^2+...+x^{10} )^5$ we still get same answer. Am wondering when will considering infinite series instead of finite one will give wrong answer? $\endgroup$ – anir Sep 26 at 10:04

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