Permutation
Let's first consider a set of \(n\) objects, which are all different. The number of all possible arrangements (permutations) is
\[ n(n-1) \cdots 1 = n! \]
Generally, if we select \(k (<n)\) objects from \(n\), the number of permuations is
\[ n(n-1)\cdot\cdot\cdot(n-k+1) = \frac{n!}{(n-k)!} = P(n, k) \]
with replacement
Previously we assumed objects were selected without replacement. If the selected object is put back into the population every time, it's called sampling with replacement. Easily, If \(k\) objects are selected from \(n\) objects with replacement, the number of permutations is
\[ n^k \]
Combination
We now consider the number of combinations of various objects when their order is immaterial. We have already known that the number of permuations of \(k\) objects from \(n\) is \(P(n, k)=\frac{n!}{(n-k)!}\), of which the permuations are repeated \(k!\) times. So the number of distinguishable arrangements (combinations) of \(k\) objects from \(n\) is
\[ \frac{n!}{k! (n-k)!} = C(n, k) \]
Let's consider \(n\) objects of which \(n_1\) are identical and of type1, \(n_2\) are identical and of type2, and \(n_m\) are identical and of type \(m\). The permutation of these \(n\) objects is \(n!\). However, the number of distinguishable arrangements is
\[ \frac{n!}{n_1! n_2! \cdots n_m!} \]
It's also called multinomial coefficients in polynomial equation \((x_1+x_2+ \cdots +x_m)^n\), where \(n\) represents the total power number of the equation, and \(n_1, n_2, \cdots n_m\) represents the power number of different variables in each expansion factors.
with replacement
As in the case of permutations we might ask how many combinations of \(k\) objects can be chosen from \(n\) objects with replacement. To calculate the value, we may image that the \(n\) distinguishable objects set out on a table. Each combination of \(k\) objects can then be made by pointing to \(k\) objects of the \(n\) objects in turn. There \(k\) equivalent selections amongst \(n\) different but re-choosable objects are strictly analogous to the placing of \(k\) indistinguishable balls into \(n\) different boxes with no restriction on the number of balls in each box.
A particular scenario in the case \(k=6, n=4\) may be symbolised as
\[ XX| \quad |XXX|X \]
There are 2 balls in the first box, none in the second box, 3 balls in the third box and 1 ball in the fourth box. So we only need to consider the combination of \(k+n-1\) objects consist of \(k\) crosses and \(n-1\) vertical lines. This is given as
\[ \frac{(k+n-1)!}{k!(n-1)!} = C(k+n-1, k) \]
Four different kinds of particle statistics
Let's consider a partical scenario. A system contains a number \(N\) of non-interacting particles, each of which can be in any of quantum states of the system. The structure of the set of quantum states is such that there exist \(R\) energy levels with corresponding energies \(E_i\) and degeneracies \(g_i\) (i.e. the \(i\)th energy level contains \(g_i\) quantum states). Find the numbers of distinct ways in which the particles can be distributed among the quantum states of the system such that the \(i\)th energy level contains \(n_i\) particles, for \(i=1,2,\cdots,R\), in the cases where the particles are
- distinguishable with no restriction on the number in each state;
- indistinguishable with no restriction on the number in each state;
- indistinguishable with a maximum of one particle in each state;
- distinguishable with a maximum of one particle in each state;
Distinguishable w/o amount restriction
Let us first consider distributing the \(N\) patricles among the \(R\) energy levels, which is analogous to placing \(N\) different balls into \(R\) boxes,
\[ \frac{N!}{\prod_{i=1}^{R} n_i!} \]
where \(n_i\) is the number of particles in each energy level. Then let's multiply the number of differnet arrangement of \(n_i\) particles in \(g_i\) degenerate quantum states in each energy level. Becasue there is no restriction on the particle number in each quantum states, each particle can reside in any of \(g_i\) degenerated quantum states. So the arrangements in this scenario is
\[ \Omega = N! \prod_{i=1}^{R} \frac{g_i^{n_i}}{n_i!} \]
Such a system of particles (for example classical gas molecules) is said to obey Maxwell-Boltzmann statistics.
Indistinguishable w/o amount restriction
If the particles are indistinguishable, there is only one arrangement in which \(n_i (i=1,2,\cdots,R)\) particles reside in \(i\)th energy level. The distinct arrangements of \(n_i\) particles in \(g_i\) degenerate states, however, is analogous to placing indistinguishable balls into different boxes. So the number of combination of particles is
\[ \Omega = \prod_{i=1}^{R} \frac{(n_i+g_i-1)!}{n_i! (g_i-1)!} \]
Such a system of particles (for example a gas of photons) is said to obey Bose-Einstein statistics.
Indistinguishable w/ amount restriction
If the number of particles in each quantum state is either 0 or 1, the arrangements of \(n_i\) particles in \(g_i\) degenerate quantum states is equal to the combination of \(n_i\) occupied and \(g_i-n_i\) unoccupied states. So the distinct arrangements of particles is given by
\[ \Omega = \prod_{i=1}^{R} \frac{g_i!}{n_i!(g_i-n_i)!} \]
Such a system of particles (for example an electron gas) is said to obey Fermi-Dirac statistics.
Distinguishable w/ amount restriction
If the particles are distinguishable, each arrangement of \(n_i\) particles in the previous scenario can be reordered in \(n_i!\) different ways. Multiplying the number of arrangements of \(N\) distinguishable particles among \(R\) energy levels, the arrangement of particles is given by
\[ \Omega = N! \prod_{i=1}^{R} \frac{g_i!}{n_i! (g_i-n_i)!} \]
Such a system of particles never occurs in nature.
References
Riley, Kenneth Franklin, et al. Mathematical methods for physics and engineering: a comprehensive guide. Cambridge university press, 2002.