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Permutations in which some are alike

Author: mahesh25

Here we learn about the permutations of the words in which some letters of the given word are alike and the rest of the words are different.

Example: MATHEMATICS, EAMCET, AIEEE,IITJEE.

If u are observe the above word say MATHEMATICS

The letters of the above word are M,A,T,H,E,M,A,T,I,C,S.

Here the letter M repeated 2 times

Here the letter A repeated 2 times

Here the letter T repeated 2 times

Here the letter H repeated 1 time

Here the letter E repeated 1 time

Here the letter I repeated 1 time

Here the letter S repeated 1 time

Here the letter C repeated 1 time

Now we find the number of linear permutations of such words.

To find the number of linear permutations of such words we use the following result

Result 1: The number of linear permutations of 'n' things in which 'p(>1)' things are alike and the rest are different is $\frac{n!}{p!}$

Result 2: The number of linear permutations of 'n' things in which there are 'p(>1)'like things of one kind ,'q(>1)' likethings of second kind, 'r'(>1) like things of which third kind and the rest are different is $\frac{n!}{p!q!r!}$ .

Example: Find the number of linear permutations of the word MATHEMATICS?

Solution: If u are observe the above word MATHEMATICS

The letters of the above word are M,A,T,H,E,M,A,T,I,C,S.

The total number of letters in the given word=11

Among the 11 letters

Here the letter M repeated 2 times

Here the letter A repeated 2 times

Here the letter T repeated 2 times

By the above result 2

Result 2: The number of linear permutations of 'n' things in which there are 'p'like things of one kind ,'q' likethings of second kind,r like things of which third kind and the rest are different is $\frac{n!}{p!q!r!}$ .

So n=11, p=2, q=2, r=2

So the number of linear permutations= $\frac{11!}{2!2!2!}$

TEST on Permutations in which some are alike and the rest are different