Answer
$W_B =1.65 \times 10^{57}$
Work Step by Step
Since configuration B has an these divisions of the molecules $n_1 = 0.6N$ and $n_2 = 0.4N$
The configuration of multiplicity for $N= 200$ is
$W_B = \frac{N!}{n_1!n_2!}$
$W_B = \frac{200!}{(0.6 \times 200)!(0.4 \times 200)!} $
$W_B = \frac{200!}{120!80!} $
$W_B =1.65 \times 10^{57}$
