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.
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
$$ \begin{align} \frac{8! -7 \times 7! + 6 * 6!}{7!-6\times6!+5\times5!} &= \frac{5!}{5!}\ \ \frac{(8\times7\times6) - 7\times(7\times6)+6\times6}{(7\times6)-6\times6+5}\\ &= \frac{7\times6+6\times6}{6+5} \\ &= \frac{13\times6}{11} \\\\ &= \boxed{\frac{78}{11}}. \end{align}$$