18. The number $\frac{20!\times 22!}{16!\times 11!}$ has $N$ prime factors, which are not necessarily distinct.
What is the value of $N(N-2)$ ?
SOLUTION
960

We can rewrite the fraction as $\frac{20!}{11!} \times \frac{22!}{16!}$.
By writing each factorial out as a product, and cancelling like factors top and bottom of each fraction, this simplifies to
$$
20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \text {. }
$$

We can count how many non-distinct prime factors each number in the above product has, and find the sum.
Taking the numbers in turn, $N=3+1+3+1+4+2+2+1+3+2+2+3+1+3+1=32$. Therefore, $N(N-2)=32 \times 30=960$.