Let f(n) be the base-10 logarithm of the product of the elements of the nth row in Pascal's triangle. Find f(10),
+605 We want:
$$\log \binom{10}{0}\binom{10}{1}\cdots \binom{10}{10}=\log \binom{10}{0}^2\binom{10}{1}^2\binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2\binom{10}{5}=\log (1\cdot 10^2\cdot 45^2\cdot 120^2\cdot 210^2\cdot 252)=\log(2^12\cdot 3^10\cdot 5^8\cdot 7^3)\approx 16.51$$
