2. Calculate the number of ways of distributing 20 different molecules among six dififerent states with the configuration 1,, 3, 5, 10, 1) Answer 20/1!0:3:5!10:1!=931170240 n0te:0!=1
2. Calculate the number of ways of distributing 20 different molecules among six different states with the configuration {1, 0, 3, 5, 10, 1}. Answer: 20! / 1! 0! 3! 5! 10! 1! = 931170240 note: 0! = 1
Stirlings Approximation: When x is large, Inx!exInx-x In xl xInx.x x124 0.000 -1.000 0.081 0.693 0.614 0.652 3.178 1.545 3.157 810.605 8.636 10.595 1630672 28.361 30.666 2042.336 39915 42.332 3074.658 72.036 74656 no,a=(2m)/2 (x+1/2)yex In Wa(NInN-N)-2(n, In n; -n:) =NInN-∑n2:lnn
Stirling’s Approximation: When x is large, ln x! x ln x - x x ln x! x ln x - x ln A 1 0.000 -1.000 0.081 2 0.693 -0.614 0.652 4 3.178 1.545 3.157 8 10.605 8.636 10.595 16 30.672 28.361 30.666 20 42.336 39.915 42.332 30 74.658 72.036 74.656 note, A = (2) 1/2 (x+1/2)x e -x ln W ( N ln N - N ) - ( ni ln ni - ni ) = N ln N - ni ln ni
The dominating configuration Imagine that N molecules distribute among two states. N, 0 N-1, 1, ., N-k, k 9··9 R1, N-1,0, N are possible configurations, and their weights are 1, N,, N!/(N-k)! k! N, 1, respectively. For instance, N=8, the weight distribution is then
The Dominating Configuration Imagine that N molecules distribute among two states. {N, 0}, {N-1, 1}, ..., {N-k, k}, ... , {1, N-1}, {0, N} are possible configurations, and their weights are 1, N, ... , N! / (N-k)! k!, ... , N, 1, respectively. For instance, N=8, the weight distribution is then
60 N=8 40 20 0 0 2345678
0 1 2 3 4 5 6 7 8 0 2 0 4 0 6 0 N = 8 W k
12000 N=16 10000 8000 6000 4000 2000 0 012345678910111213141516
0 1 2 3 4 5 6 7 8 9 1 01 11 21 31 41 51 6 0 2000 4000 6000 8000 10000 12000 N =16 W k