## Essential University Physics: Volume 1 (3rd Edition)

a) $4^n$ b) $=\frac{2n!}{(n!)(n!)}$ c) $=\frac{((N/2)!)((N/2)!)}{N!}$ d) 16.67%; 0%
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.