Given:
The word ‘MADHUBANI’
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.
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