How many permutations of the letters of the word ‘MADHUBANI’ do not begin with M but end with I?

Asked by Aaryan | 1 year ago |  76

##### Solution :-

Given:

Total number of letters = 9

A total number of arrangements of word MADHUBANI excluding I: Total letters 8. Repeating letter A, repeating twice.

The total number of arrangements that end with letter I = $$\dfrac{8! }{ 2!}$$

= $$\dfrac{8×7×6×5×4×3×2! }{ 2!}$$

= 8×7×6×5×4×3

= 20160

If the word start with ‘M’ and end with ‘I’, there are 7 places for 7 letters.

The total number of arrangements that start with ‘M’ and end with letter I = $$\dfrac{7! }{2!}$$

= 7×6×5×4×3

= 2520

The total number of arrangements that do not start with ‘M’ but end with letter I = The total number of arrangements that end with letter I – The total number of arrangements that start with ‘M’ and end with letter I

= 20160 – 2520

= 17640

Hence, a total number of arrangements of word MADHUBANI in such a way that the word is not starting with M but ends with I is 17640.

Answered by Sakshi | 1 year ago

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