Question 3.5: Determine the possible number of ways of realizing a particu...

(a) g_{i}=N_{i}=10 : We may note that \left(g_{i}-N_{i}\right) !=0 !=1. Then, from Equation (3.77), we find

W_{i}=\frac{g_{i} !}{N_{i} !\left(g_{i}-N_{i}\right) !}     (3.77)

\frac{g_{i} !}{N_{i} !\left(g_{i}-N_{i}\right) !}=\frac{10 !}{10 !}=1

(b) g_{i}=10, N_{i}=9 : We may note that \left(g_{i}-N_{i}\right) !=1 !=1. Then, we find

\frac{g_{i} !}{N_{i} !\left(g_{i}-N_{i}\right) !}=\frac{10 !}{(9 !)(1)}=\frac{(10)(9 !)}{(9 !)}=10

Comment

In part (a), we have 10 particles to be arranged in 10 quantum states. There is only one possible arrangement. Each quantum state contains one particle. In part (b), we have 9 particles to be arranged in 10 quantum states. There is one empty quantum state, and there are 10 possible positions in which that empty state may occur. Thus, there are 10 possible arrangements for this case.