 # Permutation

Permutation is an arrangement of objects in a definite order. When we look at the schedules of trains, buses and the flights we really wonder how they are scheduled according to the public’s convenience. Of course permutation is very much helpful to prepare the schedules on departure and arrival of these. Also when we come across license plates of vehicles which consists of few alphabets and digits. We can easily prepare these codes using permutations.

## Representation of Permutation:

We can represent permutation in many ways,

$\large \mathbf{P^{n}_{k}}$$\large \mathbf{_{n}P_{k}}$,

$\large \mathbf{^{n}P_{k}}$$\large \mathbf{P _{n}\, _{,k}}$,

$\large \mathbf{P(n,k)}$

## Definition of Permutation

Basically Permutation is an arrangement of objects in a particular way. While dealing with permutation one should concern about the selection as well as arrangement. In Short, Ordering is very much essential in permutations.

## Types of Permutation:

Permutation can be classified in three different categories:

• Permutation of n different objects ( when repetition is not allowed)
• Repetition, where repetition is allowed
• Permutation when the objects are not distinct (Permutation of multisets)

Let us understand both the cases of permutation in details.

## (1) Permutation of n different objects:

If n is a positive integer and r is a whole number, such that r < n, then P(n, r) represents the number of all possible arrangements or permutations of n distinct objects taken r at a time. It can also be represented as

$^{n}P_{r}$.

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

$\Rightarrow$ P(n, r) = n(n-1)(n-2)(n-3)……..(n – r +1)

$\large \Rightarrow P(n,r) = \frac{n!}{(n-r)!}$

Example: How many 3 letter words with or without meaning can be formed out of the letters of the word SWING when repetition of letters is not allowed?

Solution: Here n = 5, as the word SWING has 5 letters. Since we have to frame 3 letter words with or without meaning and without repetition, therefore total permutations possible are:

$\large \Rightarrow P(n,r) = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60$

## (2) Permutation when repetition is not allowed:

When the number of object is “n,” and we have “r” to be the selection of object, then

Choosing an object can be in n different ways (each time).

Thus the permutation of objects when repetition is allowed will be equal to,

$\large n \times n \times n \times ……(r \;\;times)$

which is given as $\large n^{r}$

Example: How many 3 letter words with or without meaning can be formed out of the letters of the word SMOKE when repetition of words is allowed?

Solution:

The number of objects, in this case, is 5, as the word SMOKE has 5 alphabets.

and r = 3, as 3 letter word has to be chosen.

Thus the permutation will be

Permutation (when repetition is allowed) = $\large 5^{3}$

$\large = 125$

## (3) Permutation of multi-sets:

Permutation of n different objects when $p_{1}$ objects among ‘n’ objects are similar, $p_{2}$ objects of the second kind are similar, $p_{3}$ objects of the third kind are similar ……… and so on, $p_{k}$ objects of the kth kind are similar and the remaining of all are of a different kind,

Thus it forms a multiset, where the permutation is given as:

$\large \mathbf{\large \frac{n!}{p_{1}!\; p_{2}!\; p_{3}…..p_{n}!}}$

## Fundamental Counting Principle

According to this principle, “If one operation can be performed in ‘m’ ways and there are n ways of performing a second operation, then the number of ways of performing the two operations together is m x n “.

This principle can be extended to the case in which the different operation be performed in m, n, p, . . . . . . ways.

In this case the number of ways of performing all the operations one after the other is m x n x p x . . . . . . . . and so on

Permutation And Combination

Permutation And Combination Class 11

Combination

## Examples on Permutation

 Example 1: In how many ways 6 children can be arranged in a line, such that (i) Two particular children of them are always together (ii) Two particular children of them are never together Solution: (i) The given condition states that, 2 students needs to be together, hence we can consider them 1. Thus, the remaining 7 gives the arrangement in 5! ways, i.e. 120. Also the two children in a line can be arranged in 2! Ways. Hence, the total number of arrangements will be, $5! \times 2! = 120 \times 2 =$ 240 ways (ii) The total number of arrangements of 6 children will be 6!, i.e. 720 ways. Out of the total arrangement, we know that, two particular children when together can be arranged in 240 ways. Therefore, total arrangement of children in which two particular children are never together will be 720 – 240 ways, i.e. 480 ways. Example 2:Consider a set having 5 elements a,b,c,d,e. In how many ways 3 elements can be selected (without repetition) out of the total number of elements. Solution: Given X = {a,b,c,d,e} 3 are to be selected. Therefore, $P_{3}^{5} = \frac{5!}{(5-3)!}$ = 60 Example 3: It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible? Solution: We are given that there are 5 men and 4 women. i.e. there are 9 positions. The even positions are , 2nd, 4th, 6th and the 8th places and These four places can be occupied by 4 women in P(4, 4) ways = 4!  = 4 . 3. 2. 1  = 24 ways The remaining 5 positions can be occupied by 5 men in P(5, 5) = 5!  = 5.4.3.2.1  = 120 ways Therefore, by the Fundamental Counting Principle,  Total number of ways of seating arrangements = 24 x 120  = 2880

## Practice Problems

Practice below listed problems:

1. How many numbers lying between 100 and 1000 can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed.
2. Seven athletes are participating in a race. In how many ways can the first three prizes be won.

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Learn about Combination and Difference between Permutation and Combination with BYJU’S-The learning App.