<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] This method is very simple :  
$\frac{1}{{9\choose r}} -\frac{1}{{10\choose r}} =\frac{{10\choose r}-{9\choose r}}{{10\choose r}{9\choose r}}=\frac{\frac{{10\choose r}-{9\choose r}}{{9\choose r}}}{{10\choose r}}=\frac{(\frac{{10\choose r}-{9\choose r}}{{9\choose r}})\frac{11*6}{11-r}}{{10\choose r}\frac{11*6}{11-r}}=\frac{11}{6{11\choose r}}\Rightarrow \\
11=(\frac{{10\choose r}-{9\choose r}}{{9\choose r}})\frac{11*6}{11-r}\Rightarrow \\
1=(\frac{{10\choose r}-{9\choose r}}{{9\choose r}})\frac{6}{11-r}\Rightarrow \\
1=(\frac{10}{10-r}-1)\frac{6}{11-r}\Rightarrow \\
1=(\frac{r}{10-r})\frac{6}{11-r}\Rightarrow \\
r^2-21r+110=6r\Rightarrow...$
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