how do I do I solve this equation

Question

(n+1)! = 110(n-1)!

I have searched online for a week now and my textbook does not explain how it got the answer: n=10.

Answer 1

Write out what those factorials mean:

$$1\cdot 2\cdot 3\cdots (n-2)(n-1)(n)(n+1) = 110 \cdot 1\cdot 2\cdot 3\cdots (n-2)(n-1).$$

What can you cancel from both sides?

Answer 2

$$(n+1)!=110(n-1)!$$ $$\frac{(n+1)!}{(n-1)!}=110$$ $$\frac{(n+1)(n)(n-1)(n-2)...(3)(2)(1)}{(n-1)(n-2)...(3)(2)(1)}=110$$ $$(n+1)(n)=110$$

You could expand this into a quadratic and solve or simply see that the only two consecutive positive integers whose product is 110 are 10 and 11. So $n=10, n+1=11, 10\cdot11=110$.