# Permutations

The number of ways of arranging*n*distinct things in a row is given by

*n*! (

*n*factorial).

The number of ways of arranging *r* objects taken at a time from *n* objects is called permutations of *n* objects by taking *r *objects at a time. Mathematically, we write it as ^{n}*P*_{r} and it is defined as

**Note: ^{n}P**

_{0}= 1

^{n}P_{1}=

**and**

*n***=**

^{n}P_{n}**!**

*n*

^{9}

*P*

_{4}.

*, we have*

^{n}P_{r }The number of ways of arranging 5 different objects is 5!

Now, we have

*F*,

*A*,

*T*and (HER). So, total 4 objects to arrange. This can be done in 4! ways.

The letters HER can then be arranged within the bundle in 3! ways.

Hence, the total words formed = 4! × 3! = 144.

3 even numbers may be placed in these 4 odd places in

^{4}

*P*

_{3}= 24 ways.

Now, 4 places are vacant and 4 odd numbers have to be placed. This can be done in 4! ways

= 4 × 3 × 2 × 1 = 24 ways.

Thus, the total number of required arrangements = 24 × 24 = 576.

When the number is of three digits, 0 cannot be the first digit or in the hundredth place. Also, since the numbers have to be 200 onwards, 1 cannot lie in the hundredth place. So, it can be filled by 2, 3, 4, 5, 6 or 7, that is, in 6 ways. Now, for the next two digits, that is, the ten’s and one’s place, there is no restriction. They can be filled out of remaining 7 digits in

^{7}

*P*

_{2}ways.

Total number of ways

When the number is of four digits, thousand place can be filled by 1, 2, 3 or 4, that is, in 4 ways. The other 3 places can be filled out of remaining 7 digits in

^{7}

*P*

_{3}ways.

Total number of ways = 4 ×

^{7}

*P*

_{3}= 840.

The numbers can be of three or four digits. Hence, the required number of ways = 252 + 840 = 1092 ways.

**Permutation of**

# Permutations of Like/Similar Things

*n*things taking all at a time, out of which

*p*things are alike of one kind and

*q*things are alike of another kind is given by ways.

Similarly, if there are three groups of similar objects, then *p* alike of one kind, *q* alike of second kind and *r* alike of a third kind.

So, no. of permutations ways

Number of 7 digit numbers

The numbers also include the numbers which have 0 to their extreme left position. These numbers should be eliminated.

Such numbers are Thus, the required number of numbers = 420 – 60 = 360.

# Circular Permutations

Permutations of objects that are considered to be lying along the circumference of a circle are called circular permutations.In circular permutations, if anticlockwise and clockwise order arrangements are considered distinct, then the number of circular permutations is (*n –1*)!*.*

Generally, this is the case when permutations are calculated for people sitting around a table. It is also referred to as “no object should have the same neighbour in any two arrangements.”

If anticlockwise and clockwise order arrangements are not distinct, then the number of circular permutations of *n* things are

Generally, in objects like flowers in a garland or beads in a necklace, we can’t distinguish clockwise and anti-clockwise arrangements.

But in clockwise and anti-clockwise arrangements, each person will have the same neighbours.

Thus, the required number of ways is

# Restricted Permutations

In many arrangements, there may be a number of restrictions. In such cases, we need to arrange or select the objects or persons as per the restrictions imposed.- Number of permutations of
*n*distinct objects when a particular object is not taken in any arrangement is^{(n – 1)}*P*_{r}_{ }.*n*objects leaving us with (*n*– 1) objects. Now, we know that permutations of (*n*– 1) objects taken*r*at a time is^{(n – 1)}*P*._{r}

ExampleHow many 3 digit numbers can be formed using the digits 1, 4, 7, 2 and 6 such that none of the numbers shall contain the digit 2?SolutionGiven that the digit 2 must not be included in any arrangement. Then, we are left with 4 digits.^{4}*P*_{3}= 24 ways. - Number of permutations of
*n*distinct objects when a particular object is always included in any arrangement is if that particular object can take any position.

ExampleFind the number of ways in which the alphabets of the word “BASIC” can be arranged taking three alphabets at a time that begin with “*B*”.SolutionGiven that*B*is included in all arrangements and its position is fixed.