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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 nPr and it is defined as

 

Description: 58819.png 
 

Note:  nP0 = 1 nP1 = n and nPn = n!

 

Example
Find the value of 9P4.
Solution
Using the equation for nPr , we have
Description: 62297.png 

 
Example
Find the number of ways in which the alphabets of the word “CHAIR” can be arranged.
Solution
There are 5 different letters in CHAIR.
The number of ways of arranging 5 different objects is 5! Description: 58834.png 

 

Example
How may four-letter words can be ma the word “TRIANGLE”?
Solution
The word triangle consists of 8 different letters. We need four-letter words. Number of permutations of 8 objects choosing 4 at a time is
Description: 58845.png 

 
Example
How many words can be formed using the letters of the word FATHER such that HER occur together?
Solution
The 3 letters HER must occur together. So, they can be treated as 1 object bundled together.
Now, we have FAT 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.

 
Example
In how many ways can the numbers 1, 2, 3, 4, 5, 6, 7 be arranged such that the even numbers may appear in odd places?
Solution
There are totally 7 digits given to us, of which 2, 4, 6 are even. In 7 places, there are 4 odd places where 3 even numbers have to be placed.
3 even numbers may be placed in these 4 odd places in 4P3 = 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.
 
Example
How many numbers between 200 and 5000 can be formed using all the digits from 0 to 7. No digit must be repeated.
Solution
The required numbers are either of three digits or four digits. We have 0 to 7, total eight digits to be used.
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 in7P2 ways.
Total number of ways Description: 58860.png
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 7P3 ways.
Total number of ways = 4 × 7P3 = 840.
The numbers can be of three or four digits. Hence, the required number of ways = 252 + 840 = 1092 ways.

Permutations of Like/Similar Things

Permutation of 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 Description: 58866.png 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 Description: 58872.png ways
 

Example
How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2 and 4?
Solution
Out of the given 7 digits, 2 occurs 3 times and 4 occurs 2 times.
Number of 7 digit numbers Description: 58878.png 
The numbers also include the numbers which have 0 to their extreme left position. These numbers should be eliminated.
Such numbers are Description: 58885.png 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 Description: 58916.png
 

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

Example
In how many ways can 7 people sit around a table so that all shall not have the same neighbours in any two arrangements?
Solution
Persons can sit around a table in (7 – 1) ! = 6 ! ways.
But in clockwise and anti-clockwise arrangements, each person will have the same neighbours.
Thus, the required number of ways is Description: 58922.png 

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.

  1. Number of permutations of n distinct objects when a particular object is not taken in any arrangement is (n – 1)Pr .
     
    When one object is never taken, we can remove it from the set of 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)Pr.

    Example
    How 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?
    Solution
    Given that the digit 2 must not be included in any arrangement. Then, we are left with 4 digits.
     
    Hence, the total number of arrangements possible are 4P3 = 24 ways.
  2. Number of permutations of n distinct objects when a particular object is always included in any arrangement is Description: 58933.png if that particular object can take any position.
     
    Number of permutations will be Description: 58939.png if the position of that particular object is fixed.

    Example
    Find 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”.
    Solution
    Given that B is included in all arrangements and its position is fixed.
     
    Total no. of arrangement Description: 58945.png 





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