Permutation Formula

The permutation formula is used to find the different number of arrangements that can be formed by taking r things from the n available things. Permutations are useful to form different words, number arrangements, seating arrangements, and for all the situations involving different arrangements. As per the permutation formula, the permutation of 'r' objects taken from 'n' objects is equal to the factorial of n divided by the factorial of difference of n and r.

What is Permutation Formula?

In the below formula of permutations, the total number of things are 'n' and 'r' number of things can be selected and arranged. Before applying the permutation formula, we need to clearly know the concept of factorial. The factorial of any number is the product of the consecutive numbers starting from 1, and till that number. i.e., n! = 1 × 2 × 3 × 4 × .......× n.

Permutation Formula

Below is the permutation formula. This is used to find the number of ways of selecting and arranging 'r' different things from 'n' different things:

• ⁿP_r (or) P(n, r) = (n!) / (n - r)!

Here:

• 'n' is the total number of things.
• 'r' is the number of things to be selected and then arranged.

Formula 1: Factorial of a natural number n.

n! = 1 × 2 × 3 × 4 × .......× n

Formula 2: Permutation Formula or NPR formula for r things taken from n things.

ⁿP_r = (n!) / (n - r)!

Formula 3: The relationship between permutations and combinations for r things taken from n things.

ⁿP_r = r! × ⁿC_r

Derivation of Permutations Formula

Since a permutation involves selecting r distinct items without replacement from n items and order is important, by the fundamental counting principle, we have

P (n, r) = n . (n-1) . (n-2) . (n-3)…… (n-(r-1)) ways.

This can be written as:

P (n, r) = n.(n-1).(n-2). (n-3) …. (n-r+1) ... (1)

Multiplying and Dividing (1) by (n-r) (n-r-1) (n-r-2)........... 3. 2. 1, we get

P (n, r) = [ n.(n-1).(n-2). (n-3) …. (n-r+1)[(n-r) (n-r-1) (n-r-2)-------- 3. 2. 1 ] / [ (n-r) (n-r-1) (n-r-2)------3. 2. 1]

P (n, r) = (n!) / (n - r)!

Hence, the permutations formula is derived.

Different Permutations Formulas

There are five different types of permutations formulas. Let us learn each of them one by one along with examples.

Permutations Formula WITHOUT Repetition

The permutations formula used when 'r' things from 'n' things have to be arranged without repetitions is nothing but the nPr formula which we have already seen. i.e.,

• ⁿP_r (or) P(n, r) = (n!) / (n - r)!

Example: Find the number 3 letter words that can be formed from the letters a, b, c, d, and e in which the letters should not be repeated.

Solution:

The number of letters available is, n = 5.

The number of letters in each word is, r = 3.

Since there should be no repetition of letters,

the possible number of words is, ⁵P₃ = (5!) / (5 - 3)! = 5! / 2! = 120/2 = 60.

Permutations Formula WITH Repetition

The permutations formula used when 'r' things from 'n' things have to be arranged with repetitions is just n^r. This is because each of the 'r' things can be selected in 'n' different ways, thus givining n×n×n× .... ×n (r times) = n^r.

Example: Find the number 3 letter words that can be formed from the letters a, b, c, d, and e in which the letters are allowed to be repeated.

Solution:

The number of letters available isn, n = 5.

The number of letters in each word is, r = 3.

Since there can be the repetition of letters,

the possible number of words is, 5³ = 125.

Permutations Formula Taken All at a Time

The number of ways of arranging 'n' different things among themselves is nothing but arranging 'n' things out of 'n' things and is given by:

ⁿP_n = n! / (n-n)! = n!/0! = n!/1 = n!

Thus, we can arrange 'n' different things among themselves in n! ways.

Example: There are 5 different books in a bookshelf. In how many ways we can arrange them?

Solution:

5 books can be arranged in 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.

Permutations Formula with Same Sets of Data

We have just seen that the number of ways of arranging 'n' different things is n!. What if all of n things are NOT different and some of them are same? Say, among 'n' things. 's₁' objects are of one type, 's₂' objects belong to the second type, ... , 's_n' objects belong to the n^th type then the number of possible arrangements is n! / (s₁! × s₂! × ... × s_n!).

Example: Find the number of arrangements of the letters of the word PETER.

Solution:

The number of letters in the given word = 5

Among them, 'E' is repeated 2 times.

So the possible number of arrangements = 5! / 2! = (120)/2 = 60.

Circular Permutations Formula

All the previous formulas refer to the arrangements in a line. But the number of ways of arranging 'n' different number of things in a circle is (n - 1)!.

Example: Find the number of ways of arranging 10 people around a round table.

Solution:

The required number of ways = (10-1)! = 9!.

Let us see the application of the permutation formula in the following solved examples.

Examples Using Permutation Formula

Example 1: Find the number of ways in which a three-digit code can be formed, using the numbers from the digits 4, 5, 7, 8, 9.

Solution:

The given digits are 4, 5, 7, 8, 9.
Total number of digits = n = 5
The number of digits in the code = r = 3
Applying the permutation formula we have:
The number of ways of forming a 3 digit code = ⁵P₃
= (5!) / (5 - 3)!
= 5!/2!
= (5 × 4 × 3 × 2 × 1)/(2 × 1)
= 5 × 4 × 3
= 60

Answer: Hence, there are 60 ways of forming a three-digit code.

Example 2: How many different words with or without meaning, can be formed using any 4 letters from a word containing 10 different letters?

Solution:

Total number of letters = n = 10
Number of letters to form the new word = r = 4
Using the permutations formula we have:
Total number of 4 letter words which can be formed = ¹⁰P₄
¹⁰P₄ = (10!) / (10-4)!
= 10!/6!
= (10 × 9 × 8 × 7 × 6!)/6!
= 10 × 9 × 8 × 7
= 5040

Answer: Hence a total of 5040 four-letter words can be formed.