Problems in Permutations

Here are more explaination about Permutations. Examples and solutions for you :)

Definition

" Permutations are the different ways in which a collection of items can be arranged " 

For example:

The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC.

Note that ABC and CBA are not same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.
The number of ways in which n things can be arranged, taken all at a time, n P n = n!, called ‘n factorial.’

Factorial Formula:

Factorial of a number n is defined as the product of all the numbers from n to 1.

For example, the factorial of 5, 5! = 5*4*3*2*1 = 120.
Therefore, the number of ways in which the 3 letters can be arranged, taken all a time, is 3! = 3*2*1 = 6 ways.
Number of permutations of n things, taken r at a time, denoted by:

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

For example:

The different ways in which the 3 letters, taken 2 at a time, can be arranged is 3!/(3-2)! = 3!/1! = 6 ways.

Important Permutation Formulas:

1! = 1
0! = 1

Let us take a look at some examples:

Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’.

Solution :

‘CHAIR’ contains 5 letters.

Therefore, the number of words that can be formed with these 5 letters = 5! = 5*4*3*2*1 = 120.
Problem 2: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘INDIA’.

Solution :

The word ‘INDIA’ contains 5 letters and ‘I’ comes twice.
When a letter occurs more than once in a word, we divide the factorial of the number of all letters in the word by the number of occurrences of each letter.
Therefore, the number of words formed by ‘INDIA’ = 5!/2! = 60.

Problem 3: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘SWIMMING?

Solution :

The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice.
Therefore, the number of words formed by this word = 8! / (2!*2!) = 10080.

Problem 4: How many different words can be formed with the letters of the word ‘SUPER’ such that the vowels always come together?

Solution :

The word ‘SUPER’ contains 5 letters.
In order to find the number of permutations that can be formed where the two vowels U and E come together.

In these cases, we group the letters that should come together and consider that group as one letter.
So, the letters are S,P,R, (UE). Now the number of words are 4.

Therefore, the number of ways in which 4 letters can be arranged is 4!

In U and E, the number of ways in which U and E can be arranged is 2!

Hence, the total number of ways in which the letters of the ‘SUPER’ can be arranged such that vowels are always together are 4! * 2! = 48 ways.

Problem 5: Find the number of different words that can be formed with the letters of the word ‘BUTTER’ so that the vowels are always together.

Solution :

The word ‘BUTTER’ contains 6 letters.
The letters U and E should always come together. So the letters are B, T, T, R, (UE).
Number of ways in which the letters above can be arranged = 5!/2! = 60 (since the letter ‘T’ is repeated twice).

Number of ways in which U and E can be arranged = 2! = 2 ways

Therefore, total number of permutations possible = 60*2 = 120 ways.

Problem 6: Find the number of permutations of the letters of the word ‘REMAINS’ such that the vowels always occur in odd places.

Solution :

The word ‘REMAINS’ has 7 letters.
There are 4 consonants and 3 vowels in it.
Writing in the following way makes it easier to solve these type of questions.
(1) (2) (3) (4) (5) (6) (7)

No. of ways 3 vowels can occur in 4 different places = 4 P 3 = 24 ways.

After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = 4 P 4 = 4! = 24 ways.
Therefore, total number of permutations possible = 24*24 = 576 ways.

** More or less hoping that I can help you to understand more about PERMUTATION.