Answer
a) $4^n$
b) $=\frac{2n!}{(n!)(n!)}$
c) $=\frac{((N/2)!)((N/2)!)}{N!}$
d) 16.67%; 0%
Work Step by Step
a) Using the equation for the number of microstates, we find:
$=4^n$
b) We find:
$=\frac{2n!}{(n!)(n!)}$
c) We find that the ratio is:
$\frac{\frac{1}{2^{n/2}}}{\frac{\frac{2n!}{(n!)(n!)}}{\frac{1}{2^{n/2}}}}$
In this case, we must plug in $N!$ for $2n!$, so we find:
$\frac{\frac{1}{2^{n/2}}}{\frac{\frac{N!}{(n!)(n!)}}{\frac{1}{2^{n/2}}}}$
$=\frac{(n!)(n!)}{N!}$
$=\frac{((N/2)!)((N/2)!)}{N!}$
d. We find:
$=\frac{((4/2)!)((4/2)!)}{4!}=.1667=\fbox{16.67%}$
$=\frac{((100/2)!)((100/2)!)}{100!}\approx0$
*Note, this is above $10^{-28}$, which is about 0.
