MATH 109
Factorials and Permutations
Given a set of N distinct objects, a permutation is an arrangement of the entire set in
order without repeats. There are N ! ways to permute the entire set. The value N ! is
called “ N factorial” and is computed by
N ! = N ! (N " 1) ! (N " 2) ! . . . ! 1 .
By convention, 0! = 1. Also 1! = 1. Then 2! = 2 ! 1 = 2 and 3! = 3 ! 2 ! 1 = 6.
Likewise, 4! = 4 ! 3 ! 2 ! 1 = 24 and 5! = 5 ! 4 ! 3 ! 2 ! 1 = 120. Your calculator should
have a built-in ! button so that you can just enter 5! rather than multiplying all the
terms out.
Example 1. A compact disc has 10 songs. The random play feature will play all of these
songs in an unknown permutation (i.e., in order, without repeats).
(a) How many possible permutations are there of these 10 songs?
(b) If you select only your 8 favorite tracks and then use the random play, then how
many possible arrangements will there be of these 8 songs?
(c) If you only have time to hear 4 songs on random play, then how many possibilities
are there for a playing of 4 different songs from the entire CD?
Solution. (a) Because all 10 songs are to be arranged in order without repeats (i.e.,
permuted), there are 10! = 3,628,800 possibilities.
(b) Now only 8 songs are to be permuted, so there are 8! = 40,320 possibilities.
(c) Now 4 songs chosen from a set of 10 are to be listed in order without repeats. So
now there are 10 ! 9 ! 8 ! 7 = 5,040 possibilities.
Note that the value in (c) also is given by
10 ! 9 ! 8 ! 7 ! 6 ! 5 ! 4 ! 3 ! 2 ! 1 10!
=
.
6 ! 5 ! 4 ! 3 ! 2 !1
6!
The 10! in the numerator comes from the total number of 10 songs. The 6! in the
denominator comes from number of unused songs in the list.
Exercises
1. Use your calculator to compute the following:
(a) 7!
(b) 13!
(c)
20!
12!
(d)
27!
21!
2. There are 11 players on the Lady Toppers with 6 seniors. (a) How many ways are
there to order the introduction of all 11 players? (b) During Senior Night, how many
ways would there be to introduce just the 6 seniors? (c) From the whole team, how
many ways are there to choose 5 players ordered as C, PF, F, SG, PG?
Permutations of a Portion of a Set
Suppose a set has N distinct objects and we wish to make a list of k of these objects (in
order without repeats). For example, from a group of 32 eligible club members, we
need 3 new officers for Vice-President, Secretary, and Treasurer. How many choices are
possible?
We are listing 3 without repeats from a group of 32, so there are 32 ! 31 ! 30 =
29,760 possible choices. (In reality though, we cut the number of choices down by using
32!
nominations.) Note that the number of choices also can be computed by
; but in
29!
this case, it is easier to use 32 ! 31 ! 30 . However, if we were arranging a larger portion
of the set, then it would be more convenient to use the factorial notation.
Example 2. In a psychological word association test, a computer will randomly pick a
letter from the alphabet (A – Z) without repeating letters. The subject will have to say
the first word coming to mind that starts with that letter. If the test goes on for 16
letters, then how many possibilities are there for the list of letters?
Solution. Because letters are not being repeated, we compute the number of choices by
26 ! 25 ! 24 ! . . . ! until we have multiplied 16 terms together. It would be much easier
26!
to use
≈ 1.111363 !10 20 (which is a lot of possibilities). The 26! comes from the total
10!
set of 26 letters, and 10! comes from the number of 10 unused letters in the list.
The Formula for P(N , k)
The expression P(N, k ) , also written n P r on some calculators, is computed by
P(N, k ) =
N!
.
(N ! k )!
You must use the difference in the denominator. For example,
P(14, 6) =
14!
14!
=
= 2,162,160.
(14 ! 6)!
8!
You can divide these two factorials 14! ÷ 8! on your calculator; however, your
calculator may have a built-in n P r button. Then compute P(14, 6) by entering the
command 14 n P r 6, which still gives 2,162,160.
The formula for P(N, k ) gives the number of ways to permute a group of k objects
selected from larger group of N objects. Thus when permuting 16 letters from the
entire alphabet of 26, there are P(26, 16) possibilities which is calculated by
P(26, 16) =
26!
≈ 1.111363 !10 20 .
10!
Exercises
3. Use your calculator to compute the following:
(a) P(16, 7)
(b) P(52, 5)
4. Suppose you have one Scrabble piece for each of the 26 letters. How many ways are
there to permute six of these pieces?
5. You have 12 players on your intramural softball team. (a) How many ways are there
to make a batting order of 10 players? (b) How many ways are there to arrange the
entire team in order?
6. A new lottery game called “Match 5” is being created. The player must pick five
numbers chosen from 1 – 20 without repeating numbers. To win, the player must
match the winning numbers in order. How many ways are there to pick the five
numbers and what is the probability of winning on one play?
Answers
1. (a) 5,040
(b) 6,227,020,800
2. (a) 11! = 39,916,800
3. (a) 57,657,600
(c) 5,079,110,400
(b) 6! = 720
(d) 213,127,200
(c) 11 ! 10 ! 9 ! 8 ! 7 = 11! ÷ 6! = 55440
(b) 311,875,200
4. P(26, 6) = 165,765,600
5. (a) P(12, 10) = 239,500,800
(b) 12! = 479,001,600
6. P(20, 5) = 1,860,480 ways; the probability of winning is p =
1
.
1,860, 480