Permutation Formula

Permutations are arrangements of objects or elements in a specific order. The permutation formula calculates the number of ways in which these arrangements can be formed. It is an essential concept in combinatorics, the branch of mathematics concerned with counting and arranging objects.

The permutation formula computes the number of permutations of r objects chosen from a set of n distinct objects, where r is less than or equal to n. It is denoted by P(n,r) or nPr and is given by:

P(n,r)=n×(n−1)×(n−2)×…×(n−r+1)

Understanding permutations and their formulas is crucial for solving problems involving arrangements and selections in various mathematical and real-world contexts.

Permutation Meaning

Permutation is the arrangement of things or elements in a particular order. Permutations are used in mathematics, notably combinatorics, to quantify the number of possible arrangements. The term “permutation” originated from the Latin word “permūtātiō,” which means “exchange” or “change.”

Permutations depend on the sequence in which the elements are arranged. For example, the arrangement “ABC” differs from “ACB” or “CAB.”

Permutations involve the arrangement of various objects or elements. Each element is distinct and identifiable from the others.

Depending on the context, permutations may or may not allow for element repetition.

What is Permutation Formula?

The permutation formula calculates the number of ways in which objects or elements can be arranged in a specific order. It is used to determine the number of permutations of r objects chosen from a set of n distinct objects, where r is less than or equal to n. The permutation formula is denoted by P(n,r) or nPr.

P(n,r)=n!/(n−r)!

Where:

• n represents the total number of distinct objects available for selection.
• r represents the number of objects to be chosen or arranged.
• n! denotes the factorial of n, which is the product of all positive integers up to n.
• (n−r)! represents the factorial of n−r, which is the product of all positive integers from 11 to n−r

Derivation of Permutation Formula 

To derive the permutation formula, let’s consider a set of n distinct objects from which we want to arrange r objects in a specific order. We can think of this process as a series of r choices, where we choose one object from the set at a time, with each choice reducing the number of available objects by one.

The number of choices for the first object is n, as there are 𝑛n distinct objects in the set. After making the first choice, the number of available choices for the second object decreases to n−1, as one object has already been chosen. Similarly, for the third object, there are n−2 choices, and so on, until we have made r choices.

By the principle of multiplication, the total number of arrangements of r objects chosen from n distinct objects is the product of the number of choices at each step.

So, the number of arrangements can be expressed as:

P(n,r)=n×(n−1)×(n−2)×…×(n−r+1)

This product represents the number of permutations of r objects chosen from n distinct objects. However, this expression can be simplified using factorial notation.

the product n×(n−1)×(n−2)×…×(n−r+1) is equivalent to n factorial divided by (n−r)!. This is because:

​n×(n−1)×(n−2)×…×(n−r+1)= n!/(n−r)!

​permutation formula is derived as:

P(n,r)=n!/(n−r)!

Different Permutations Formulas

The formulas for permutations depend on whether repetition is allowed or not. Here are the different permutation formulas:

Permutations Formula WITHOUT Repetition 

If repetition is not allowed, and there are n distinct objects to choose from and r positions to fill (where r≤n), the formula for permutations is given by P(n,r)=

n!/(n−r)!

​Example: If we have 5 distinct letters (A, B, C, D, E) and want to arrange them in a 3-letter word without repetition, there are

𝑃(5,3)=5!/(5−3)! = 5!/2! = (5×4×3×2×1)/(2×1) = 60

Permutations Formula WITH Repetition

​If repetition of elements is allowed, and there are n distinct objects to choose from and r positions to fill, the formula for permutations is simply n^r. This is because for each of the r positions, there are n choices.

Example: If we have 3 distinct letters (A, B, C) and want to arrange them in a 3-letter word with repetition allowed, there are 3³ = 27 possible permutations.

Permutations Formula Taken All at a Time 

The permutations formula for “taken all at a time” refers to the number of ways to arrange r objects chosen from a set of n distinct objects, where r is equal to n. In other words, it calculates the number of permutations of all n objects without repetition.

The formula for permutations “taken all at a time” is: P(n,n)=n!

n! denotes the factorial of n, which is the product of all positive integers up to n.

Suppose we have a set of 4 distinct letters: A, B, C, and D. We want to find the number of ways to arrange all 4 letters in a specific order.

Using the permutations formula for “taken all at a time,” we have:

P(4,4)=4!

Now, let’s calculate the factorial:

4!=4×3×2×1=24

So, there are 24 different permutations of all 4 letters without repetition.

Each permutation represents a unique arrangement of all 4 letters in a specific order, such as ABCD, ACBD, BACD, etc.

Permutations Formula with Same Sets of Data 

When we talk about permutations of the same sets of data, we usually mean permutations in which some of the elements in the set are similar, resulting in repeated permutations. In such circumstances, the total number of permutations is lowered since arrangements with identical elements are deemed identical. To compute permutations using the same sets of data, we alter the permutation formula by dividing out the redundancies created by repeated items. We employ factorials to account for these repeats.

Let’s denote:

n as the total number of objects.

n1 ,n2,…,nk as the counts of each type of identical object.

The formula for permutations with the same sets of data is:

P = n!/n₁​!⋅n₂​!⋅…⋅n_k​!

Suppose we want to arrange the letters of the word “MISSISSIPPI.”

Here, we have:

n=11 (total number of letters)

n₁​=1 (count of ‘M’)

n_2​=4 (count of ‘I’)

n₃​=4 (count of ‘S’)

n₄​=2 (count of ‘P’)

Using the formula:

P=11!/1!⋅4!⋅4!⋅2!

P=39916800/1152

P≈34650

Circular Permutations Formula

The formula for circular permutations is derived from the linear permutations formula. Suppose we have n distinct objects arranged in a circle. The number of distinct circular permutations is given by:

P_circular​=(n−1)!​/2

​Suppose we have 5 distinct objects arranged in a circle. Using the circular permutations formula:

Pcircular​=(5−1)!​/2 =4!​/2=24​/2=12

Examples Using Permutation Formula 

Example 1: How many ways can you arrange 3 letters from the word “BRIGHT”?

Solution:

• $n=6$ (since there are 6 letters in “BRIGHT”)
• $r=3$ (we are arranging 3 letters)

P(6,3)=6!/(6−3)!​=6!/3! = 720/6 = 120

Example 2: In how many ways can you select and arrange 4 books from a shelf of 10 books?

Solution:

• $n=10$
• r=4

P(10,4) = 10!​/(10−4)! = 10!/6!

10!=10×9×8×7×6!

Since the 6! terms cancel out, we have:​

$P(10,4)=10\times 9\times 8\times 7=5040$

There are 5040 ways to select and arrange 4 books from 10.

Example 3: How many ways can a 2-member committee be formed and arranged from a group of 5 people?

Solution:

• $n=5$
• $r=2$

P(5,2)= 5!​/(5−2)! = 5!/3!

Since, 5!=5×4×3!

Since the

3! terms cancel out, we have:

P(5,2)=5×4=20

There are 20 ways to form and arrange a 2-member committee from 5 people.​