Permutation – SS2 Further Mathematics Lesson Note (Second Term)

A permutation is an arrangement of objects in which the order matters. The number of ways of arranging n different objects is n! (n factorial). When only r of the n objects are arranged at a time, the number of permutations is nPr = n!/(n-r)!. For example, the letters p, q and r can be arranged in 6 ways.

Behavioural objectives

By the end of the lesson, students should be able to:

1. State what a permutation is.
2. Explain n! and use it to find the number of arrangements of n different objects.
3. Simplify expressions involving factorials.
4. Use the formula nPr = n!/(n-r)! to evaluate permutations of n objects taken r at a time.
5. Solve simple word problems on arrangements and seating.

What is a permutation?

Permutation is defined as the number of arrangements of objects. The different orders of arrangement are important.

Example: Find the number of ways of arranging the letters p, q, r.

The arrangements are pqr, prq, qpr, qrp, rpq, rqp. The number of ways is 6.

Similarly, for 4 different letters the number of arrangements is 24.

How do we find the number of arrangements of n different objects?

In general, the number of different arrangements of n different objects is equal to n! (n factorial).

n! = n × (n-1) × (n-2) × … × 3 × 2 × 1, and 0! = 1.

Example 1: simplifying factorials

Simplify the following:

1. 5!/3!
2. 7!/4!

Solution

1. 5!/3! = (5 × 4 × 3!)/3! = 5 × 4 = 20
2. 7!/4! = (7 × 6 × 5 × 4!)/4! = 7 × 6 × 5 = 210

Example 2: arranging the letters of a word

Find the number of ways of arranging the letters of the word MACHINE.

Solution: There are seven different letters in the word MACHINE, therefore the number of permutations is 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 ways.

How do we arrange n objects taken r at a time?

Sometimes we want the number of ways of arranging only some of the objects, for example 2 letters taken from a 4-lettered word. The number of permutations of n objects taken r at a time is written nPr, where:

nPr = n!/(n-r)!

Example 3: evaluating nPr

Evaluate: (a) 8P3 (b) 11P9

Solution:

(a) 8P3 = 8!/(8-3)! = 8!/5! = (8 × 7 × 6 × 5!)/5! = 8 × 7 × 6 = 336

(b) 11P9 = 11!/(11-9)! = 11!/2! = 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 19958400

Example 4: seating arrangement

In how many ways can three people be seated on eight seats in a row?

Solution:

The 1st seat can be occupied by any of the 8 people, so there are 8 ways. The 2nd seat can then be occupied in 7 ways, and the 3rd seat in 6 ways.

Hence, the number of ways = 8 × 7 × 6 = 336 ways.

Alternatively, n = 8, r = 3:

nPr = n!/(n-r)! = 8!/(8-3)! = 8!/5! = 8 × 7 × 6 = 336 ways.

Evaluation

1. What is a permutation?
2. Simplify 6!/4!.
3. Find the number of ways of arranging the letters of the word CHAIRS.
4. Evaluate 7P3.

Assignment

1. Evaluate 9P4.
2. In how many ways can four people be seated on six seats in a row?
3. Find the number of ways of arranging the letters of the word LAGOS.