# What is the easiest way to solve factorial division questions like these?

I have a question. I am new here so I am sorry if this question is bad or vague. I'm having some issues solving this one factorial question: shown below:

$$8! -7 * 7! + 6 * 6! \over 7!-6*6!+5*5!$$

I know the answer is $$78/11$$. I don't understand the way to solve this though. The solution said to split $$8!$$ into $$8 * 7 * 6!$$ and $$7!$$ into $$7 * 6!$$ for the numbers in the numerator. Then you get $$8 * 7 -7 * 7 + 6 * 6!$$. I know that they removed the $$6!$$ from the numerator and multiplied the $$6!$$ to $$13$$ which is the simplified expression $$8 * 7 -7 * 7 + 6$$. The part I don't understand is why they did that. In the original equation, there are 3 $$6!$$ (in the numerator's original equation). Why do they multiply $$6!$$? Thanks in advance.

• Not really following what you wrote. Every term in the numerator is divisible by $6!$ and every term in the denominator is divisible by $5!$ Since $\frac {6!}{5!}=6$, that gets us to $6\times \frac {8\times 7-7\times 7 +6}{7\times 6-6\times 6 +5}$ which is easy to work with. Does that answer your question? – lulu Jan 14 '20 at 0:35

A factorial is just a long falling product. $$3! = 3\times 2\times 1,\, 4! = 4\times3! = 4\times 3 \times 2 \times 1$$ and so on. In this specific question, note that every term in the numerator and denominator of the fraction has a factorial. So we should factor those products. How? Well, if $$a > b$$, then $$b\times (b-1)\times\ldots\times1$$ should appear in the $$a!$$ product, just like $$3!$$ appears in $$4!$$. So we pick the least factorial appearing, which is $$5!$$. So
\begin{align} \frac{8! -7 \times 7! + 6 * 6!}{7!-6\times6!+5\times5!} &= \frac{5!}{5!}\ \ \frac{(8\times7\times6) - 7\times(7\times6)+6\times6}{(7\times6)-6\times6+5}\\ &= \frac{7\times6+6\times6}{6+5} \\ &= \frac{13\times6}{11} \\\\ &= \boxed{\frac{78}{11}}. \end{align}