Compute $\frac{6! + 7!}{5!}$
I notice that both the numerator and the denominator have factorials, which are products of consecutive integers. I wonder if I can simplify them by canceling out some common factors. I recall that $n! = n \times (n-1) \times \cdots \times 1$, so I can write $6! = 6 \times 5!$ and $7! = 7 \times 6!$. Substituting these expressions into the fraction, I get $\frac{6 \times 5! + 7 \times 6 \times 5!}{5!}$. I see that I can factor out $5!$ from both terms in the numerator, so I get $\frac{5!(6 + 7 \times 6)}{5!}$. Now I can cancel out $5!$ from the numerator and the denominator, leaving me with $6 + 7 \times 6$. I can use the order of operations to simplify this expression by doing the multiplication first, then the addition. I get $6 + 42 = 48$. 