Math 141: Business Mathematics I
Fall 2015
§6.4 Permutations and Combinations
Instructor: Yeong-Chyuan Chung
Outline
• Factorial notation
• Permutations - arranging objects
• Combinations - selecting objects
Factorial notation
Recall that in §6.3, when we were applying the multiplication principle, we often (but not
always) got expressions such as 3 · 2 · 1 or 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1. The factorial notation
allows us to write such expressions more concisely. The first expression would be written as
3! and the second expression would be written as 10!.
Given any positive integer n, we write n! (read “n factorial”) to mean n·(n−1)·(n−2) · · · 2·1.
By convention, we also define 0! = 1.
Example. 1! = 1, 2! = 2 · 1 = 2, 5! = 5 · 4 · 3 · 2 · 1 = 120.
Using a calculator to compute factorials
1. Enter the number.
2. Press MATH, and move to the PRB menu. Select 4:!. Press Enter.
Example. How many ways are there to seat 8 people in a row of 8 seats?
1
§6.4 Permutations and Combinations
2
Permutations - arranging objects
Given a set of objects, a permutation of the set is an arrangement of the objects in a
definite order.
If the objects in the set are all distinct, then as we saw in the previous section (and the
previous example), the number of possible arrangements can be found by applying the multiplication principle.
If we are using up all the objects in the set, then the answer will simply be given by some
factorial. In the previous example, there were 8 seats and there were 8 people to arrange so
the multiplication principle tells us that the number of possible arrangements is 8 · 7 · 6 · · · 1,
which is simply 8!. But what if there were only 5 seats? In this case, the multiplication
principle tells us that the number of possible arrangements is 8 · 7 · 6 · 5 · 4. Since we are
not multiplying all the way down to 1, this expression is not the factorial of any number
but nevertheless it is related to the factorials of some numbers - observe that it is equal to
8!
8 · 7 · 6···4 · 3 · 2 · 1
= .
3·2·1
3!
In general, if we have n distinct objects and we want to arrange r of them, then the number
of possible arrangements is given by
n!
.
P (n, r) =
(n − r)!
Sometimes, we also say that it is the number of permutations of n distinct objects taken r
at a time.
As a special case, if we want to arrange all n objects (meaning r = n), then the number of
n!
n!
=
= n! (since 0! = 1 by definition).
possible arrangements would be P (n, n) =
(n − n)!
0!
Using a calculator to compute P (n, r)
1. Enter the number n.
2. Press MATH, and move to the PRB menu. Select 2:nPr.
3. Enter the number r. Press Enter.
Example. In how many ways can we select 8 people from a group of 60 and arrange them
in 8 chairs?
Example (Exercise 26 in the text). How many three-digit numbers can be formed by using
the numerals in the set {3, 2, 7, 9} if repetition is not allowed?
§6.4 Permutations and Combinations
3
We have discussed the case where all the objects are distinct. When some of the objects are
identical, we will get fewer possible arrangements because interchanging the identical objects
results in the same arrangement. Consider the following example.
Suppose we have 2 blue balls and 1 red ball. If we pretend for a moment that the two blue
balls are different, say by labeling them as B1 and B2 , then from what we have discussed
previously, we would get 3! possible arrangements. We can list all the possible arrangements
as follows:
RB1 B2 RB2 B1 B1 RB2 B2 RB1 B1 B2 R B2 B1 R
Now we remember that the two blue balls were actually indistinguishable and remove their
labels. Then we get the following list:
RBB
RBB
BRB
BRB
BBR BBR
So we see that there are actually only 3 possible arrangements.
To get the correct number of possible arrangements, we need to account for over-counting as
a result of interchanging identical objects. In the example above, the correct number, 3, was
3!
where the 3! in the numerator comes from pretending that every ball is distinct,
actually
2!
and the 2! in the denominator comes from the number of possible arrangements of the two
identical blue balls.
In general, if we have n objects in which n1 objects are alike and of one kind, n2 objects
are alike and of another kind, ..., and nm objects are alike and of yet another kind, so that
n1 + n2 + · · · + nm = n, then the number of permutations of these n objects (taken n at a
time) is given by
n!
.
n1 !n2 ! · · · nm !
Example. Suppose we have 2 identical red marbles, 3 identical green marbles, and 1 blue
marble. If we want to line the marbles up in a row, how many distinguishable arrangements
of the 6 marbles are there?
Example (Exercise 40 in the text). Find the number of distinguishable permutations that
can be formed from the letters of the word PHILIPPINES.
Example (Exercise 44 in the text). How many different signals can be made by hoisting two
yellow flags, four green flags, and three red flags on a ship’s mast at the same time?
§6.4 Permutations and Combinations
4
Combinations - selecting objects
We have seen that permutations involve arrangements of objects so the order matters in
these situations. In some cases, we are only interested in the number of ways of selecting a
certain number of objects (in other words, a subset) from a collection without any regard to
the order in which they are selected. These subsets are called combinations.
The number of combinations of n objects taken r at a time is given by
C(n, r) =
n!
.
(n − r)!r!
Notice that this formula looks similar to that for the number of permutations of n distinct
objects taken r at a time. The extra term r! in the denominator comes from disregarding
the order in which the r objects are arranged.
Using a calculator to compute C(n, r)
1. Enter the number n.
2. Press MATH, and move to the PRB menu. Select 3:nCr.
3. Enter the number r. Press Enter.
Example. In how many ways can I pick a group of 3 from a class of 60 students?
Example (Exercise 38 in the text). In how many ways can an investor select four mutual
funds for his investment portfolio from a recommended list of eight mutual funds?
§6.4 Permutations and Combinations
5
Examples - distinguishing between permutations and
combinations; dealing with restrictions
Example (Exercise 48 in the text). A group of five students studying for a bar exam has
formed a study group. Each member of the group will be responsible for preparing a study
outline for one of five courses. In how many different ways can the five courses be assigned
to the members of the group?
Example (Exercise 52 in the text). L.A. Wedding Caterers offers a wedding reception buffet.
Suppose a menu is planned around four different salads, six entrees, six side dishes, and seven
desserts. There are eight different choices of salads, ten different choices of entrees, eight
different choices of side dishes, and ten different choices of desserts. How many menus are
possible?
Example (Exercise 54 in the text). A company car that has a seating capacity of six is to
be used by six employees who have formed a car pool. If only four of these employees can
drive, how many possible seating arrangements are there for the group?
Example (Exercise 56 in the text). The town of Carson employs 15 police officers. On a
typical day, 6 of the officers are to be assigned to duty in patrol cars, 4 are assigned to the
foot patrol, and the remaining 5 are assigned to duty at the station. How many different job
configurations are there?
§6.4 Permutations and Combinations
6
Example (Exercise 58 in the text). In how many ways can four married couples attending
a concert be seated in a row of eight seats if:
(a) There are no restrictions?
(b) Each married couple is seated together?
(c) The members of each sex are seated together?
Example (Exercise 68 in the text). In how many ways can a subcommittee of four be chosen
from a Senate committee of five Democrats and four Republicans if:
(a) All members are eligible?
(b) The subcommittee must consist of two Republicans and two Democrats?