# How to find the value of $r$ such that $\frac{1}{\binom{9}{r}} - \frac{1}{\binom{10}{r}} = \frac{11}{6\binom{11}{r}}$?

Find the value of $$r$$:

$$\frac{1}{\dbinom{9}{r}} - \frac{1}{\dbinom{10}{r}} = \frac{11}{6\dbinom{11}{r}}$$

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.

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Since $$\binom{n}{k} = \begin{cases} \dfrac{n!}{k!(n - k)!} & \text{if 0 \leq k \leq n}\\[2 mm] 0 & \text{if k > n} \end{cases}$$ we require that $$k \leq 9$$ since otherwise we would be dividing by $$0$$.

\begin{align*} \frac{1}{\binom{9}{k}} - \frac{1}{\binom{10}{k}} & = \frac{11}{6\binom{11}{k}}\\ \frac{1}{\dfrac{9!}{k!(9 - k)!}} - \frac{1}{\dfrac{10!}{k!(10 - k)!}} & = \frac{11}{6 \cdot \dfrac{11!}{k!(11 - k)!}} && \text{by definition of \binom{n}{k}}\\ \frac{k!(9 - k)!}{9!} - \frac{k!(10 - k)!}{10!} & = \frac{11k!(11 - k)!}{6 \cdot 11!} && \text{division is multiplication by the reciprocal}\\ \frac{10k!(9 - k)!}{10!} - \frac{k!(10 - k)!}{10!} & = \frac{k!(11 - k)!}{6 \cdot 10!} && \text{form common denominator on LHS, cancel on RHS}\\ 6[10k!(9 - k)! - k!(10 - k)!] & = k!(11 - k)! && \text{multiply by 6 \cdot 10!}\\ 6[10 - (10 - k)] & = (11 - k)(10 - k) && \text{divide by k!(9 - k)!}\\ 6k & = 110 - 21k + k^2 && \text{simplify}\\ 0 & = k^2 - 27k + 110 && \text{set quadratic equal to zero} \end{align*} When you solve the quadratic equation, keep in mind the restriction that $$k \leq 9$$.

Hopefully you know the formula $$\dbinom{n}{k} = \frac{n!}{(n-k)!k!}$$.

So your equation becomes: $$\frac{r!(9-r)!}{9!} - \frac{r!(10-r)!}{10!} = \frac{11\times r!(11-r)!}{6\times 11!}$$

$$\frac{r!(9-r)!}{9!} - \frac{r!(9-r)!(10-r)}{9!\times 10} = \frac{11\times r!(9-r)!(10-r)(11-r)}{6\times 9!\times10\times11}$$

Divide both sides by $$\frac{r!(9-r)!}{9!}$$:

$$1-\frac{10-r}{10} = \frac{11\times (10-r)(11-r)}{6\times 10\times11}$$

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