Question

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

Answer 1

Hint: Firstly we will calculate total number of possible arrangements that end with letter ‘I’. Once we have that we will find the total number of arrangements when the word starts with ‘M’ and ends with ‘I’, Then we just have to subtract both values to get the desired result.

Complete answer:
It is given that we have to find a permutation for the word ‘MADHUBANI’ which has a total 9 letters.
As the total number of arrangements of word MADHUBANI excluding I which means Total 8 letters.
Here letter A is repeating twice in this word.
Here total number of arrangements that end with the letter ‘I’ are = \[\dfrac{{8!}}{{2!}}\] .
Opening Factorials for further simplification.
\[ \Rightarrow \dfrac{{\left[ {8{\rm{ }} \times {\rm{ }}7{\rm{ }} \times {\rm{ }}6{\rm{ }} \times {\rm{ }}5{\rm{ }} \times {\rm{ }}4{\rm{ }} \times {\rm{ }}3{\rm{ }} \times {\rm{ }}2!} \right]}}{{2!}}\]
Cancelling \[2!\] from denominator and numerator.
So, we get \[ \Rightarrow 8{\rm{ }} \times {\rm{ }}7{\rm{ }} \times {\rm{ }}6{\rm{ }} \times {\rm{ }}5{\rm{ }} \times {\rm{ }}4{\rm{ }} \times {\rm{ }}3 = 20160\]
Hence, the total number of arrangements that end with letter I are 20160
Now we will calculate the number of arrangements when the word starts with ‘M’ and ends with ‘I’ then there would be 7 vacant places which should be filled by the remaining 7 letters.
So, the total number of arrangements that start with ‘M’ and end with letter ‘I’ = \[\dfrac{{7!}}{{2!}}\]
\[ \Rightarrow \dfrac{{\left[ {{\rm{ }}7{\rm{ }} \times {\rm{ }}6{\rm{ }} \times {\rm{ }}5{\rm{ }} \times {\rm{ }}4{\rm{ }} \times {\rm{ }}3{\rm{ }} \times {\rm{ }}2!} \right]}}{{2!}}\]
Cancelling \[2!\] from denominator and numerator.
So, we get \[ \Rightarrow 7{\rm{ }} \times {\rm{ }}6{\rm{ }} \times {\rm{ }}5{\rm{ }} \times {\rm{ }}4{\rm{ }} \times {\rm{ }}3 = 2520\]
Hence, the total number of arrangements that start with ‘M’ and end with letter I are 2520.
Therefore the total number of arrangements that do not start with ‘M’ but end with letter I can be calculated by deducting the total number of arrangements that start with ‘M’ and end with letter I from the total number of arrangements that end with letter I .
So, we get \[ \Rightarrow {\rm{ }}20160{\rm{ }}-{\rm{ }}2520 = 17640\] .

Hence, the total number of arrangements of the letters of the word MADHUBANI do not begin with ‘M’ but end with ‘I’ are 17640.

Note: To solve these types of questions, we must remember that we have to divide the permutation with the factorial of total repeating letters. Like in this question A was repeating twice so we divided permutation by \[2!\] .