Answer
$\frac{W_A}{W_B} = 4.16 \times 10^{16} $
Work Step by Step
Taking the equation from (b), when $N = 100$ the $n_m$ should be divided with the same amount. And because 100 is not evenly divisible by 3, put 34 molecules into one of the three box parts of configuration A and 33 in each of the other two parts.
$\frac{W_A}{W_B} = \frac{(N/2!)(N/2!)}{(N/3!)(N/3!)(N/3!)}$
$\frac{W_A}{W_B} = \frac{(50!)(50!)}{(33!)(33!)(34!)}$
$\frac{W_A}{W_B} = 4.16 \times 10^{16} $
