Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PROBABILITY

Another useful result that may be derived using the binomial coefficients is the

number of ways in whichndistinguishable objects can be divided intompiles,

withniobjects in theith pile,i=1, 2 ,...,m(the ordering of objects within each

pile being unimportant). This may be straightforwardly calculated as follows. We

may choose then 1 objects in the first pile from the originalnobjects innCn 1 ways.

Then 2 objects in the second pile can then be chosen from then−n 1 remaining

objects inn−n^1 Cn 2 ways, etc. We may continue in this fashion until we reach the

(m−1)th pile, which may be formed inn−n^1 −···−nm−^2 Cnm− 1 ways. The remaining

objects then form themth pile and so can only be ‘chosen’ in one way. Thus the

total number of ways of dividing the originalnobjects intompiles is given by

the product

N=nCn 1 n−n^1 Cn 2 ···n−n^1 −···−nm−^2 Cnm− 1

=

n!

n 1 !(n−n 1 )!

(n−n 1 )!

n 2 !(n−n 1 −n 2 )!

···

(n−n 1 −n 2 −···−nm− 2 )!

nm− 1 !(n−n 1 −n 2 −···−nm− 2 −nm− 1 )!

=

n!

n 1 !(n−n 1 )!

(n−n 1 )!

n 2 !(n−n 1 −n 2 )!

···

(n−n 1 −n 2 −···−nm− 2 )!

nm− 1 !nm!

=

n!

n 1 !n 2 !···nm!

. (30.35)

These numbers are calledmultinomial coefficientssince (30.35) is the coefficient of

xn 11 xn 22 ···xnmmin the multinomial expansion of (x 1 +x 2 +···+xm)n, i.e. for positive

integern

(x 1 +x 2 +···+xm)n=

∑

n 1 ,n 2 ,... ,nm

n 1 +n 2 +···+nm=n

n!

n 1 !n 2 !···nm!

xn 11 xn 22 ···xnmm.

For the casem=2,n 1 =k,n 2 =n−k, (30.35) reduces to the binomial coefficient
nCk. Furthermore, we note that the multinomial coefficient (30.35) is identical to

the expression (30.32) for the number of distinguishable permutations ofnobjects,

niof which are identical and of typei(fori=1, 2 ,...,mandn 1 +n 2 +···+nm=n).

A few moments’ thought should convince the reader that the two expressions

(30.35) and (30.32) must be identical.

In the card game of bridge, each of four players is dealt 13 cards from a full pack of 52.

What is the probability that each player is dealt an ace?

From (30.35), the total number of distinct bridge dealings is 52!/(13!13!13!13!). However,
the number of ways in which the four aces can be distributed with one in each hand is
4!/(1!1!1!1!) = 4!; the remaining 48 cards can then be dealt out in 48!/(12!12!12!12!)
ways. Thus the probability that each player receives an ace is

4!

48!

(12!)^4

(13!)^4

52!

=

24(13)^4

(49)(50)(51)(52)

=0. 105 .

As in the case of permutations we might ask how many combinations ofk

objects can be chosen fromnwith replacement(repetition). To calculate this, we