# 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.

• Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. When you pose a question here, you should show what you have attempted in addition to explaining where you got stuck. That helps users identify any mistakes you may have made and write answers appropriate to your skill level. – N. F. Taussig Sep 25 '18 at 0:56

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.