<s>[INST] $\frac{1}{{9\choose r}} -\frac{1}{{10\choose r}} = \frac{11}{6{11\choose r}}$. Is there a way to find $r$ without using algebra? 
$$\frac{1}{\dbinom 9r} -\frac{1}{{\dbinom{10}r}} = \frac{11}{6\times \dbinom{11}r}$$

I guess directly applying algebra for this problem would be enough. But are there any simpler and prettier approaches to finding $r$ ?
 [/INST] Well, you can use Pascal's rule, or whatever it's called:
$$ \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} $$
to rewrite the left-hand side as
$$ \frac{\binom{10}{r}-\binom{9}{r}}{\binom{10}{r}\binom{9}{r}} = \frac{\binom{9}{r-1}}{\binom{10}{r}\binom{9}{r}}. $$
Multiplying up gives
$$ \frac{11}{6}\binom{10}{r}\binom{9}{r} = \binom{9}{r-1}\binom{11}{r} $$
Our next trick is another well-known property of binomial coefficients, namely
$$ \binom{n}{k+1}=\frac{n-k}{k+1}\binom{n}{k} $$
Use this to rewrite
$$ \binom{9}{r} = \frac{10-r}{r}\binom{9}{r-1}, $$
so 
$$ \frac{11}{6}\binom{10}{r}\binom{9}{r} = 11\binom{10}{r}\frac{10-r}{r}\binom{9}{r-1} = \binom{9}{r-1}\binom{11}{r} \\
\frac{11}{6}\binom{10}{r}\frac{10-r}{r} = \binom{11}{r} $$
Now we have another binomial identity,
$$ \binom{n+1}{k}=\frac{n+1}{n+1-k}\binom{n}{k}, $$
so
$$ \binom{11}{6r} = \frac{11}{11-r}\binom{10}{r}, $$
reducing the equation to
$$ \frac{1}{11-r} = \frac{10-r}{6r}, $$
which is the quadratic you get normally. This is really not inherently different, but does avoid getting factorials out.
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