3.2 Permutations & Combinations
3.2
1
Permutations & Combinations
Definition: A permutation of n distinct objects is an arrangement or ordering of the n objects.
The number of such arrangements is denoted n Pn .
An r-permutation of n distinct objects is an arrangement using r of the n objects. The number
of possible r-permutations is denoted n Pr .
An r-combination of n distinct objects is an unordered selection or
of r out of the n
µ subset
¶
n
objects. The number of such selections is denoted by n Cr or simply
.
r
Note: Using the multiplication rule, it can be shown that n Pn = n(n − 1)(n − 2) . . . 3.2.1 = n!
and that
n
Pr = n(n − 1)(n − 2) . . . (n − r + 1)
·
(n − r)(n − r − 1)(n − r − 2) . . . 2.1
= n(n − 1)(n − 2) . . . (n − r + 1)
(n − r)(n − r − 1)(n − r − 2) . . . 2.1
n(n − 1)(n − 2) . . . (n − r + 1)(n − r)(n − r − 1)(n − r − 2) . . . 2.1
=
(n − r)(n − r − 1)(n − r − 2) . . . 2.1
n!
=
(n − r)!
¸
Also, all r-permutations of n objects can be generated by first picking any r-combination of the
n objects and then arranging these r objects in any order. Thus, the number of r-permutations
n
Pr
n!
is given by n Pr =n Cr r Pr so n Cr = r =
.
Pr
(n − r)! r!
Examples:
1. Suppose there are eight athletes in a race. The winner receives a gold medal, the secondplace finisher receives a silver medal and the third-place finisher receives a bronze medal.
How many different ways can these medals be awarded?
The number of different ways the medals can be awarded is the number of 3-permutations
8!
of 8 elements: 8 P3 = 8.7.6 =
= 336.
(8 − 3)!
2. A salesman has to visit eight different cities. He must begin his trip in one particular city
but can visit the other cities in any order he wishes. In how many different ways can he
make his trips?
The number of different ways the salesman can make his trip or the number of possible
paths between the cities is the number of permutations of 7 elements: 7 P7 = 7! = 5040.
3. How many ways are there to pick a 5-person basketball team from 10 possible players
(a) without restriction?
(b) if the weakest and the strongest players must be on the team?
10!
= 252,
5! 5!
8!
= 56.
(b) 8 C3 =
5! 3!
(a)
10
C5 =
c Sinéad Breen, 2004
°
3.2 Permutations & Combinations
2
4. Suppose a committee of k people is to be chosen from a set/group of 7 females and 4
males. How many ways are there to form the committee if
(a) it consists of 3 females and one is to be named chairperson, one treasurer and a third
secretary?
(b) it consists of 3 females with no distinction of duties?
(c) it consists of 3 females and 2 males with no distinction of duties?
(d) it consists of 4 people at least 2 of whom are female (with no distinction of duties)?
(a) As members of the committee have distinct duties, the number of possible committees
is the number of arrangements using 3 of the 7 females: 7 P3 = 7.6.5 = 210.
(b) As members of the committee do not have distinct duties, the number of possible committees is the number of unordered selections using 3 of the 7 females:
7
P3
210
7
C3 =
=
= 35.
3!
6
(c) Applying the multiplication rule to compose the number
µ of¶subsets of 3 females with
4!
the number of subsets of 2 males gives 7 C3 4 C2 = 35
= 35(6) = 210.
2! 2!
(d) A committee of 4 people with at least 2 females has 2 females, 3 females or 4 females.
Thus the number of such committees possible is given by
7
C2 4 C2 + 7 C3 4 C1 + 7 C4 4 C0 = 301.
5. How many different 8-digit binary sequences are there with exactly 6 ones?
Exactly 6 of the 8 digits are ones: there are 8 C6 = 28 ways in which the positions for the
ones can be chosen and the remaining positions can be filled in just one way with zeroes.
(Alternatively, there are 8 C2 = 28 ways in which two positions can be chosen for the zeroes
and the remaining positions are filled with ones).
Remarks:
1. The numbers n Cr are frequently called binomial coefficients because of their appearance
as coefficients in the binomial expansion (x + y)n .
2. n Cr = n Cn−r as n Cr =
n!
n!
n!
and n Cn−r =
=
.
(n − r)! r!
(n − (n − r))! (n − r)!
r! (n − r)!
There is often more than one way to look at a particular problem:
Example: How many ways are there to arrange the letters in the word SYSTEMS?
(i) Three of the seven positions in any arrangement will be filled with S: there are 7 C3 ways
in which this can be done. The remaining four positions can be filled with Y, T, E, M in
7!
4! = 7.6.5.4 = 840.
4.3.2.1 ways. Thus, the overall number of arrangements is 7 C3 .(4!) =
3! 4!
c Sinéad Breen, 2004
°
3.2 Permutations & Combinations
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(ii) There would be 7 P7 = 7! ways to arrange the seven letters if they were all distinct, but
there are three Ss so we must divide by the number of ways these can be arranged which
7!
is
= 7.6.5.4 = 840.
3!
( 7 P7 counts the arrangement ‘SYSTEMS’ six times for instance: this can be seen by
labelling the Ss
S1 YS2 TEMS3 S1 YS3 TEMS2 S2 YS1 TEMS3 S2 YS3 TEMS1 S3 YS1 TEMS2 S3 YS2 TEMS1 )
(iii) Another approach to answering this question is discussed in the next section.
c Sinéad Breen, 2004
°