Why Read This Book? - Index of

4.9 Combinations and Partitions 151
who ordered what. In how many different ways could he set these 4 dishes down
among the 6 people at the table?
4.9 Combinations and Partitions
In this section, we build on this result to determine the number of ways we may
select m out of n objects, where we do not distinguish between different arrangements.
Then we generalize this to determine how many ways we may partition a
given set into distinguishable subsets of given sizes.
4.9.1 Combinations
Given a set of n objects and an integer 0 ≤ m ≤ n, we can calculate the number of
ways we may choose m of the n objects, with no order taken into consideration.
This is precisely the same as asking how many distinct subsets containing m objects
may be formed from a set with n objects. We call this the number of combinations
of � n objects taken m at a time, and we denote it C(n, m). An alternate notation is
n�
m , which is read “n choose m.”
A natural way to derive the formula for �n �
m is with an equivalence relation.
Let S be the set of all the P(n, m) permutations of the n objects taken m at a
time. Define two permutations in S to be equivalent provided they contain precisely
the same elements of the set, regardless of the order of these elements.
Clearly, this is an equivalence relation, so that S is partitioned into equivalence
classes, where any given class contains all the possible arrangements of a particular
m-element subset of the n objects. The number of equivalence classes is therefore
C(n, m), the expression whose formula we are trying to determine. Notice that
each equivalence class contains precisely m! elements. Since |S| = P(n, m), we
have that m!×C(n, m) = P(n, m), so that
C(n, m) =
We have therefore proved the following.
� �
n
=
m
P(n, m)
m! =
n!
m!(n − m)!
(4.25)
Theorem 4.9.1 Suppose A is a finite set of cardinality n, and let 0 ≤ m ≤ n.
Then the number of subsets of A of cardinality m is given by Eq. (4.25).
Notice that C(n, m) = C(n, n − m), so the number of ways to choose m out of n
objects is the same as the number of ways to choose n − m of the objects. Notice
also that C(n, 0) = 1 is meaningful in that there is precisely one zero-element
subset of a given set.