For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set of
n
n
objects and we want to choose
r
r
objects from the set in order, we write
P(n,r).
P(n,r).
Another way to write this is
n
P
r
,
n
P
r
,
a notation commonly seen on computers and calculators. To calculate
P(n,r),
P(n,r),
we begin by finding
n!,
n!,
the number of ways to line up all
n
n
objects. We then divide by
(
n−r
)!
(
n−r
)!
to cancel out the
(
n−r
)
(
n−r
)
items that we do not wish to line up.

Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is
6×5×4=120.
6×5×4=120.
Using factorials, we get the same result.

6!
3!
=
6·5·4·3!
3!
=6·5·4=120
6!
3!
=
6·5·4·3!
3!
=6·5·4=120

There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows.

P(n,r)=
n!
(n−r)!
P(n,r)=
n!
(n−r)!

Note that the formula stills works if we are choosing __all__
n
n
objects and placing them in order. In that case we would be dividing by
(
n−n
)!
(
n−n
)!
or
0!,
0!,
which we said earlier is equal to 1. So the number of permutations of
n
n
objects taken
n
n
at a time is
n!
1
n!
1
or just
n!.
n!.

Given
n
n
distinct objects, the number of ways to select
r
r
objects from the set in order is

P(n,r)=
n!
(n−r)!
P(n,r)=
n!
(n−r)!

(4)
*Given a word problem, evaluate the possible permutations.*

- Identify
n
n
from the given information.
- Identify
r
r
from the given information.
- Replace
n
n
and
r
r
in the formula with the given values.
- Evaluate.

A professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions?

Substitute
n=12
n=12
and
r=9
r=9
into the permutation formula and simplify.

P(n,r)=
n!
(n−r)!
P(12,9)=
12!
(12−9)!
=
12!
3!
=79,833,600
P(n,r)=
n!
(n−r)!
P(12,9)=
12!
(12−9)!
=
12!
3!
=79,833,600

There are 79,833,600 possible permutations of exam questions!

*Could we have solved Example 4 using the Multiplication Principle?*

*Yes. We could have multiplied*
15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4
15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4
*to find the same answer*.

A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find the following.

How many ways can the 7 actors line up?

P(7,7)=5,040
P(7,7)=5,040

How many ways can 5 of the 7 actors be chosen to line up?

P(7,5)=2,520
P(7,5)=2,520

AnalysisNote that in part c, we found there were 9! ways for 9 people to line up. The number of permutations of
n
n
distinct objects can always be found by
n ! .
n ! .