Permutations and combinations — an application to Statistical Mechanics

In Statistical Mechanics, one  encounters the situation of putting k particles into r distinct energy levels. The particles can thus be considered as discrete objects and the different energy levels as distinct boxes or cells. Three different situations are obtained by making  three different assumptions. These are:

a) Maxwell-Boltzmann: Here the particles are all distinct and any number of particles can be put into any of the r boxes. The number of possibilities are $r^{k}$ as given by the following theorem:

Let M be a multi set consisting of r distinct objects, each with infinite multiplicity. Then, the total number of d-permutations of M is $r^{d}$.

b) Bose-Einstein: Here the particles are all identical and any number of particles can be put in any one of the r boxes. The number of possibilities is $k+r-1 \choose k$ as given by the following theorem:

The following sets are in bijective correspondence:

i) The set of all increasing sequences of  length  k on an ordered set with r elements.

ii) The set of all the ways of putting k identical objects into  r distinct boxes.

iii) The set of all the k-combinations of a multi-set with r distinct elements.

c) Fermi-Dirac: Here the particles are all identical  but no  box can hold more than one particle. The number of possibilities is $r \choose k$.

More later…

Nalin Pithwa

A cute problem on permutations and combinations

Here, is a problem on counting. Read it and crack it before you compare your solution with the one I present !!!

In a tennis tournament, there are $2n$ participants. In the  first round of  the tournament, each participant plays just once, so there are n games, each occupying a pair of  players. Show that the pairing for the first round can be arranged in exactly

$1 x 3 x 5 x 7 x 9 \ldots(2n-1)$

different ways.

Solution.

Call the required number of pairings of $2n$ players $P_{n}$. If you are a participant, you can be matched with any one of  the other $(2n-1)$ players. Once your antagonist is chosen, there remain

$(2n-2)=2(n-1)$

players who  can be paired in $P_{n-1}$ ways. Hence,

$P_{n}=(2n-1)P_{n-1}$.