2016 AMC 10A Problem 1

Never multiply out $11!$ by hand. Instead, notice how neighboring factorials are linked.

1. Recall that $11! = 11 \cdot 10 \cdot 9 \cdots 1$, which is the same as $11 \cdot 10!$. In the same way, $10! = 10 \cdot 9!$.
2. So the top of the fraction is
$$11! - 10! = 11 \cdot 10! - 1 \cdot 10!.$$
That is "$11$ copies of $10!$ take away $1$ copy of $10!$", which leaves $10$ copies:
$$11! - 10! = 10 \cdot 10!.$$
3. Now divide by $9!$. Since $10! = 10 \cdot 9!$, dividing $10!$ by $9!$ gives $10$:
$$\frac{10 \cdot 10!}{9!} = 10 \cdot 10 = 100.$$

The answer is $\boxed{\textbf{(B)}\ 100}$.

Each factorial is the one before it times one more number, so neighboring factorials share almost everything. Pull out the common factorial first and only small numbers are left to multiply. Whenever you see factorials added, subtracted or divided, look for the largest factorial they all contain.

Split the fraction into two pieces and simplify each one:
$$\frac{11!}{9!} - \frac{10!}{9!} = (11 \cdot 10) - 10 = 110 - 10 = 100.$$

Both terms of the numerator contain $10!$, since $11! = 11 \cdot 10!$. Factor it out:
$$11! - 10! = 11 \cdot 10! - 10! = 10 \cdot 10!.$$

Now divide by $9!$. Because $10! = 10 \cdot 9!$, we get
$$\frac{10 \cdot 10!}{9!} = 10 \cdot 10 = 100.$$

The answer is $\boxed{\textbf{(B)}\ 100}$.