Section 4.4 Permutations and Combinations
273
P(n, 0) =
n!
(n − 0)!
=
n!
n!
= 1
This formula can be interpreted as saying that there is only one ordered arrangement of zero objects—the empty set.
P(n, 1) =
n!
(n − 1)!
= n
This formula reflects the fact that there are n ordered arrangements of 1 object.
(Each arrangement consists of the 1 object, so this merely counts how many ways
to get the 1 object.)
P(n, n) =
n!
(n − n)!
=
n!
0!
= n!
This formula states that there are n! ordered arrangements of n distinct objects,
which merely reflects the multiplication principle—n choices for the first
object, n − 1 choices for the second object, and so on, with 1 choice for the nth
object.
example 47
The number of permutations of 3 objects, say a, b, and c, is given by
P(3, 3) = 3! = 3 # 2 # 1 = 6. The 6 permutations of a, b, and c are
abc, acb, bac, bca, cab, cba
Note that we could have solved Example 48 just by using the multiplication
principle—there are 8 choices for the first letter, 7 for the second, and 6 for the
third, so the answer is 8 # 7 # 6 = 336. P(n, r) simply gives us a new way to think
about the problem, as well as a compact notation.
example 48
How many three-letter words (not necessarily meaningful) can be formed from the
word “compiler” if no letters can be repeated? Here the arrangement of letters matters, and we want to know the number of permutations of 3 distinct objects taken
from 8 objects. The answer is P(8, 3) = 8!/5! = 336.
example 49
Ten athletes compete in an Olympic event. Gold, silver, and bronze medals are
awarded; in how many ways can the awards be made?
This problem is essentially the same as the one in Example 48. Order matters; given 3 winners A, B, and C, the arrangement A−gold, B−silver, C−bronze
is different than the arrangement C−gold, A−silver, B−bronze. So we want the
number of ordered arrangements of 3 objects from a pool of 10, or P(10, 3). Using
the formula for P(n, r), P(10, 3) = 10!/7! = 10 # 9 # 8 = 720.
273
P(n, 0) =
n!
(n − 0)!
=
n!
n!
= 1
This formula can be interpreted as saying that there is only one ordered arrangement of zero objects—the empty set.
P(n, 1) =
n!
(n − 1)!
= n
This formula reflects the fact that there are n ordered arrangements of 1 object.
(Each arrangement consists of the 1 object, so this merely counts how many ways
to get the 1 object.)
P(n, n) =
n!
(n − n)!
=
n!
0!
= n!
This formula states that there are n! ordered arrangements of n distinct objects,
which merely reflects the multiplication principle—n choices for the first
object, n − 1 choices for the second object, and so on, with 1 choice for the nth
object.
example 47
The number of permutations of 3 objects, say a, b, and c, is given by
P(3, 3) = 3! = 3 # 2 # 1 = 6. The 6 permutations of a, b, and c are
abc, acb, bac, bca, cab, cba
Note that we could have solved Example 48 just by using the multiplication
principle—there are 8 choices for the first letter, 7 for the second, and 6 for the
third, so the answer is 8 # 7 # 6 = 336. P(n, r) simply gives us a new way to think
about the problem, as well as a compact notation.
example 48
How many three-letter words (not necessarily meaningful) can be formed from the
word “compiler” if no letters can be repeated? Here the arrangement of letters matters, and we want to know the number of permutations of 3 distinct objects taken
from 8 objects. The answer is P(8, 3) = 8!/5! = 336.
example 49
Ten athletes compete in an Olympic event. Gold, silver, and bronze medals are
awarded; in how many ways can the awards be made?
This problem is essentially the same as the one in Example 48. Order matters; given 3 winners A, B, and C, the arrangement A−gold, B−silver, C−bronze
is different than the arrangement C−gold, A−silver, B−bronze. So we want the
number of ordered arrangements of 3 objects from a pool of 10, or P(10, 3). Using
the formula for P(n, r), P(10, 3) = 10!/7! = 10 # 9 # 8 = 720.