Question

The 120 permutations of $\text{MAHES}$ are arranged in dictionary order as if each were an ordinary 5 letter word .The last letter of $86^{th}$ word in the list is
A) $A$
B) $H$
C) $S$
D) $E$

Answer 1

Hint: Here first we will arrange the given letters into alphabetical order and then find the number of words starting with each letter in alphabetical order and stop at $86^{th}$ word to find the last letter of that word.

Complete step by step answer:
First we will arrange the given letters into alphabetical order:
Letters in alphabetical order are:- $A,E,H,M,S$
Now will find the number of words starting with $A$.
So, we will fix $A$ at the first place and then fill other places with remaining letters.

A

Now since second place can be filled in 4 ways , third place can be filled in 3 ways , fourth place can be filled in 2 ways and fifth place can be filled in 1 way
Therefore, The number of words starting with $A$ are : \[4 \times 3 \times 2 \times 1\]
                                                                            \[ = 24\] ways………………….(1)
Now we will find the number of words starting with $E$.
So, we will fix $E$ at the first place and then fill other places with remaining letters.

E

Now since second place can be filled in 4 ways , third place can be filled in 3 ways , fourth place can be filled in 2 ways and fifth place can be filled in 1 way
Therefore, The number of words starting with $E$ are : \[4 \times 3 \times 2 \times 1\]
                                                                            \[ = 24\] ways……………………….(2)
Now we will find the number of words starting with $H$.
So, we will fix $H$ at the first place and then fill other places with remaining letters.

H

Now since second place can be filled in 4 ways , third place can be filled in 3 ways , fourth place can be filled in 2 ways and fifth place can be filled in 1 way
Therefore, The number of words starting with $H$ are : \[4 \times 3 \times 2 \times 1\]
                                                                            \[ = 24\] ways……………………….(3)
Now we will find the number of words starting with $M$.
So we will fix $M$ at the first place and then fill other places with remaining letters.

M

Now since second place can be filled in 4 ways , third place can be filled in 3 ways , fourth place can be filled in 2 ways and fifth place can be filled in 1 way
Therefore, The number of words starting with $M$ are: \[4 \times 3 \times 2 \times 1\]
                                                                            \[ = 24\] ways……………………….(3)
Adding equations 1, 2, 3 and 4 we get:
\[
  24 + 24 + 24 + 24 \\
   = 96 \\
 \]
Hence we know that $86^{th}$ word starts with letter $M$
Now we will fix another letter $A$ along with $M$ and then find the number of words starting with ‘MA’

M

A

Now first and second places can be filled by only 1 way since the letters are fixed third place can be filled by 3 ways , fourth place can be filled by 2 ways and fifth place can be filled by 1 way
Therefore, Letters starting with ‘MA’ are :- \[3 \times 2 \times 1\]
                                                        \[ = 6\] ways
Similarly letter starting with ‘ME’ are”:- \[3 \times 2 \times 1\]
                                                        \[ = 6\] ways
Therefore now we have reached
\[
  72 + 6 + 6 \\
   = 84{\text{ words}} \\
 \]
Therefore, $85^{th}$ word starts with ‘MH’
And it is ‘MHAES’
Now if we fix ‘MHA’ then $86^{th}$ word would be ‘MHASE’.
Hence the last letter of the $86^{th}$ word is $E$.

Hence, option D is the correct option.

Note:
In such questions , since the words are arranged in a dictionary hence they should be in alphabetical order. Also, the number of words in which the letters can be arranged can also be calculated by using factorials.