# $\frac{x}{10!} = \frac{1}{8!} + \frac{1}{9!}$

I have a pretty simple straightforward question.

Q) Find the value of $x$ in the following: $$\frac{x}{10!} = \frac{1}{8!} + \frac{1}{9!}$$

Instinctively, I do the quickest thing I know how to do. $$\times10!\implies x = \frac{10!}{8!} + \frac{10!}{9!} = 9 \times 10 + 9 = 99$$ That seemed pretty easy and the back of my textbook says it's the right answer.

Just for fun, I do it again. Except this time, I initially multiply the equation with $9!$ instead of $10!$,

$$\times 9!\implies\frac{9!}{10!}x = \frac{9!}{8!} + \frac{9!}{9!}$$ $$\implies\frac{x}{10} = 9 + 1$$ $$\times 10\implies x = 10\times10 = 100$$ Sure, by multiplying by $10$ in the last step, I've still multiplied by $10!$.
But now, this $1\%$ variation in the answer has popped up.

So, I did what any highschooler would do in this situation;
I consulted Wolfram Alpha which told me that my $2^{\text{nd}}$ trial was correct.

Who should I believe? A computational knowledge engine or the back of my textbook?

• $$\frac{(n+1)!}{n!}=n+1\;\ldots$$ – DonAntonio Mar 1 '14 at 17:42
• Let me answer your question: you should believe your brain! – user63181 Mar 1 '14 at 17:42
• @Sami: I never thought about that, but I will keep it mind. – Nick Mar 1 '14 at 18:02
• @Don: Will recheck my calculations before I post questions again. Sir, yes, sir. – Nick Mar 1 '14 at 18:04
• Lessons learnt: (1) I am an idiot who should find his brain and (2) do not doubt Wolfram's calculations. – Nick Mar 1 '14 at 18:09

$$x = \frac{10!}{8!} + \frac{10!}{9!} = 10 \times 9 + 10 = 100$$