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Find the value of $r$:

$$frac1dbinom9r - frac1dbinom10r = frac116dbinom11r$$

I'm not sure where I should take this problem in order to isolate $r$. I seem to get many factorials that don't seem to combine.

Since
$$binomnk =
begincases
dfracn!k!(n - k)! & textif $0 leq k leq n$\[2 mm]
0 & textif $k > n$
endcases$$
we require that $k leq 9$ since otherwise we would be dividing by $0$.

beginalign*
frac1binom9k - frac1binom10k & = frac116binom11k\
frac1dfrac9!k!(9 - k)! - frac1dfrac10!k!(10 - k)! & = frac116 cdot dfrac11!k!(11 - k)! && textby definition of $binomnk$\
frack!(9 - k)!9! - frack!(10 - k)!10! & = frac11k!(11 - k)!6 cdot 11! && textdivision is multiplication by the reciprocal\
frac10k!(9 - k)!10! - frack!(10 - k)!10! & = frack!(11 - k)!6 cdot 10! && textform common denominator on LHS, cancel on RHS\
6[10k!(9 - k)! - k!(10 - k)!] & = k!(11 - k)! && textmultiply by $6 cdot 10!$\
6[10 - (10 - k)] & = (11 - k)(10 - k) && textdivide by $k!(9 - k)!$\
6k & = 110 - 21k + k^2 && textsimplify\
0 & = k^2 - 27k + 110 && textset quadratic equal to zero
endalign*
When you solve the quadratic equation, keep in mind the restriction that $k leq 9$.

Hopefully you know the formula $dbinomnk = fracn!(n-k)!k!$.

So your equation becomes:
$$fracr!(9-r)!9! - fracr!(10-r)!10! = frac11times r!(11-r)!6times 11!$$

$$fracr!(9-r)!9! - fracr!(9-r)!(10-r)9!times 10 = frac11times r!(9-r)!(10-r)(11-r)6times 9!times10times11$$

Divide both sides by $fracr!(9-r)!9!$:

$$1-frac10-r10 = frac11times (10-r)(11-r)6times 10times11$$

I hope it's now clear that this is a quadratic equation. I haven't tried to solve it, but the work should be simple.

EDIT: I totally forgot about the restriction. Taussig's answer is correct, after solving the quadratic equation you must eliminate any r>9.

Making the problem more general
$$frac1dbinomnr - frac1dbinomn+1r = frackdbinomn+2r$$ and using the same approach as N. F. Taussig in his/her answer, you should end with a quadratic equation in $r$
$$k r^2- left(k(2 n+3 )+n+2right)r+kleft( n^2+3 n+2right)=0$$ If $k >0$, there are two positive roots $left(Delta=k^2+2 k (n+2) (2 n+3)+(n+2)^2,r_1+r_2>0,r_1r_2>0right)$ and, as said in answers, you must discard the one which is > n.