The Chemistry Maths Book, Second Edition

21.6 Permutations and combinations 611

Equation (21.26) is already very accurate whenn 1 = 152 , and can be assumed to be

exact for the larger values of nin the table. The difference between (21.26) and the

‘more approximate’ (21.27) is , and this increases only very slowly with n.

The absolute error of 27.4 whenn 1 = 110

23

corresponds to the fractional error 51 × 110

− 24

inln 1 n!

An important use of Stirling’s formula is given in Example 21.12 below. We

consider first how a distribution, such as the binomial distribution, behaves when

the numbers involved are large. Figure 21.4 shows the binomial distribution (21.18)

forn 1 = 110 andn 1 = 1100. The graphs are ofP

m

2 P

n 22

againstm 2 n, with maximum value

1 whenm 2 n 1 = 1122.

They show that when nis large the properties of the distribution are dominated by

the probabilities at and close to the maximum. A measure of this relative narrowing

of the distribution curve is given by the fractional standard deviation.

EXAMPLE 21.12The Boltzmann distribution

In statistical thermodynamics, the Boltzmann distribution for a system of nparticles

with total energy Eis often derived by considering the number of ways the particles

can be distributed amongst the particle states. Let there be kavailable states with

energiesε

1

, ε

2

, =, ε

k

, and letn

1

particles have energyε

1

,n

2

have energyε

2

, and so on.

The number of ways of arranging the nparticles in this way (assuming distinguishable

particles) is then given by theorem 4, equation (21.24):

(21.28)

When nis large, as it is in a thermodynamic system, the properties of the system

are dominated by the distribution for which Whas its maximum value. This most

probable distribution is obtained by optimizing Wwith respect to the occupation

W

n

nn n

k

=

!

!!!

12

σnn=12()

1

2

ln 2 πn

..... ... .. ... .. ... .. .

... ... .. ... .. ... .. .

......... ........ ....

...

.........

.......

m/n

P

m

/P

n/ 2

0 1 / 2 1

1

..

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n= 10

..... ... .. ... .. ... .. .

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...

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m/n

P

m

/P

n/ 2

0 1 / 2 1

1

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n=100

Figure 21.4