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A probability question I was trying to solve the other reduced to this simple statement:

Find the coefficient of $x^{21}$ in the expansion of:

$$(1+x+x^2+x^3+x^4+x^5)^6$$

Obviously, I can get a computer to do this in seconds, but I would be happier if I could solve the whole problem without computer help. Is there any simple way to answer this question? I've looked at Multinomial expansions, but I can't seem to get them to work.

Thanks

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A general solution was given here. –  draks ... May 25 '12 at 15:00
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1 Answer 1

up vote 1 down vote accepted

Writing $[x^n]s(x)$ for the coefficient of $x^n$ in the expansion of the series $s(x)$, one looks for $[x^{21}]u(x)$ with $u(x)=(1-x^6)^6v(x)$ and $v(x)=(1-x)^{-6}$.

Since $(1-x^6)^6=1-6x^6+15x^{12}-20x^{18}+o(x^{21})$, $$ [x^{21}]u(x)=[x^{21}]v(x)-6[x^{15}]v(x)+15[x^{9}]v(x)-20[x^{3}]v(x). $$ Now, $5!\cdot v(x)$ is the fifth derivative of $(1-x)^{-1}$ hence $v(x)=\sum\limits_{n\geqslant0}{n+5\choose 5}x^n$ and

$$ [x^{21}]u(x)={26\choose 5}-6{20\choose 5}+15{14\choose 5}-20{8\choose 5}=1666. $$

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