3.2 Permutations & Combinations 1

3.2 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 nPn.An r-permutation of n distinct objects is an arrangement using r of the n objects. The numberof possible r-permutations is denoted nPr.An r-combination of n distinct objects is an unordered selection or subset of r out of the n

objects. The number of such selections is denoted by nCr or simply

(n

r

).

Note: Using the multiplication rule, it can be shown that nPn = n(n− 1)(n− 2) . . . 3.2.1 = n!and that

nPr = n(n− 1)(n− 2) . . . (n− r + 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− 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 then objects and then arranging these r objects in any order. Thus, the number of r-permutations

is given by nPr =n CrrPr so nCr =

nPr

rPr

=n!

(n− r)! r!.

Examples:

1. Suppose there are eight athletes in a race. The winner receives a gold medal, the second-place 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

of 8 elements: 8P3 = 8.7.6 =8!

(8− 3)!= 336.

2. A salesman has to visit eight different cities. He must begin his trip in one particular citybut can visit the other cities in any order he wishes. In how many different ways can hemake his trips?The number of different ways the salesman can make his trip or the number of possiblepaths between the cities is the number of permutations of 7 elements: 7P7 = 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?

(a) 10C5 =10!

5! 5!= 252,

(b) 8C3 =8!

5! 3!= 56.

c© Sinead 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 4males. 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 thirdsecretary?

(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 committeesis the number of arrangements using 3 of the 7 females: 7P3 = 7.6.5 = 210.

(b) As members of the committee do not have distinct duties, the number of possi-ble committees is the number of unordered selections using 3 of the 7 females:

7C3 =7P3

3!=

210

6= 35.

(c) Applying the multiplication rule to compose the number of subsets of 3 females with

the number of subsets of 2 males gives 7C34C2 = 35

(4!

2! 2!

)= 35(6) = 210.

(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

7C24C2 + 7C3

4C1 + 7C44C0 = 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 8C6 = 28 ways in which the positions for theones can be chosen and the remaining positions can be filled in just one way with zeroes.(Alternatively, there are 8C2 = 28 ways in which two positions can be chosen for the zeroesand the remaining positions are filled with ones).

Remarks:

1. The numbers nCr are frequently called binomial coefficients because of their appearanceas coefficients in the binomial expansion (x + y)n.

2. nCr = nCn−r as nCr =n!

(n− r)! r!and nCn−r =

n!

(n− (n− r))! (n− r)!=

n!

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 7C3 waysin which this can be done. The remaining four positions can be filled with Y, T, E, M in

4.3.2.1 ways. Thus, the overall number of arrangements is 7C3.(4!) =7!

3! 4!4! = 7.6.5.4 = 840.

c© Sinead Breen, 2004

3.2 Permutations & Combinations 3

(ii) There would be 7P7 = 7! ways to arrange the seven letters if they were all distinct, butthere are three Ss so we must divide by the number of ways these can be arranged which

is7!

3!= 7.6.5.4 = 840.

( 7P7 counts the arrangement ‘SYSTEMS’ six times for instance: this can be seen bylabelling the SsS1YS2TEMS3 S1YS3TEMS2 S2YS1TEMS3 S2YS3TEMS1 S3YS1TEMS2 S3YS2TEMS1 )

(iii) Another approach to answering this question is discussed in the next section.

c© Sinead Breen, 2004