Section 6.
3
Section Summary
Permutations
Combinations
Combinatorial Proofs
Permutations
Definition: A permutation of a set of distinct objects
is an ordered arrangement of these objects. An ordered
arrangement of r elements of a set is called an
r-permuation.
Example: Let S = {1,2,3}.
The ordered arrangement 3,1,2 is a permutation of S.
The ordered arrangement 3,2 is a 2-permutation of S.
The number of r-permuatations of a set with n
elements is denoted by P(n,r).
The 2-permutations of S = {1,2,3} are 1,2; 1,3; 2,1; 2,3;
3,1; and 3,2. Hence, P(3,2) = 6.
A Formula for the Number of
Permutations
Theorem 1: If n is a positive integer and r is an integer with
1 ≤ r ≤ n, then there are
P(n, r) = n(n − 1)(n − 2) ∙∙∙ (n − r + 1)
r-permutations of a set with n distinct elements.
Proof: Use the product rule. The first element can be chosen in n
ways. The second in n − 1 ways, and so on until there are
(n − ( r − 1)) ways to choose the last element.
Note that P(n,0) = 1, since there is only one way to order zero
elements.
Corollary 1: If n and r are integers with 1 ≤ r ≤ n, then
Solving Counting Problems by
Counting Permutations
Example: How many ways are there to select a first-
prize winner, a second prize winner, and a third-prize
winner from 100 different people who have entered a
contest?
Solution:
P(100,3) = 100 ∙ 99 ∙ 98 = 970,200
Solving Counting Problems by
Counting Permutations (continued)
Example: Suppose that a saleswoman has to visit eight
different cities. She must begin her trip in a specified
city, but she can visit the other seven cities in any order
she wishes. How many possible orders can the
saleswoman use when visiting these cities?
Solution: The first city is chosen, and the rest are
ordered arbitrarily. Hence the orders are:
7! = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 5040
If she wants to find the tour with the shortest path that
visits all the cities, she must consider 5040 paths!
Solving Counting Problems by
Counting Permutations (continued)
Example: How many permutations of the letters
ABCDEFGH contain the string ABC ?
Solution: We solve this problem by counting the
permutations of six objects, ABC, D, E, F, G, and H.
6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720
Combinations
Definition: An r-combination of elements of a set is an
unordered selection of r elements from the set. Thus, an
r-combination is simply a subset of the set with r elements.
The number of r-combinations of a set with n distinct
elements is denoted by C(n, r). The notation is also
used and is called a binomial coefficient.
Example: Let S be the set {a, b, c, d}. Then {a, c, d} is a 3-
combination from S. It is the same as {d, c, a} since the
order listed does not matter.
C(4,2) = 6 because the 2-combinations of {a, b, c, d} are the
six subsets {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, and {c, d}.
Combinations
Theorem 2: The number of r-combinations of a set
with n elements, where n ≥ r ≥ 0, equals
Proof: By the product rule P(n, r) = C(n,r) ∙ P(r,r).
Therefore,
Combinations
Example: How many poker hands of five cards can be dealt
from a standard deck of 52 cards? Also, how many ways are
there to select 47 cards from a deck of 52 cards?
Solution: Since the order in which the cards are dealt does
not matter, the number of five card hands is:
The different ways to select 47 cards from 52 is
This is a special case of a general result. →
Combinations
Corollary 2: Let n and r be nonnegative integers with
r ≤ n. Then C(n, r) = C(n, n − r).
Proof: From Theorem 2, it follows that
and
Hence, C(n, r) = C(n, n − r).
This result can be proved without using algebraic manipulation. →
Combinations
Example: How many ways are there to select five players from a
10-member tennis team to make a trip to a match at another
school.
Solution: By Theorem 2, the number of combinations is
Example: A group of 6 people have been trained as astronauts to
go on the first mission to Mars. How many ways are there to
select 3 crews of two people to go on this mission?
Solution:
C(6,2) ⋅ C(4,2) ⋅ C(2,2)
= 15
(3,3)
