JEE MAIN - Mathematics (2025 - 22nd January Morning Shift - No. 8)

From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ' M ', is :
6084
5148
14950
4356

Explanation

First, note that we are choosing 5 distinct letters (in strictly increasing alphabetical order) such that the middle (third) letter is ‘M’. Symbolically, if we denote the chosen letters as:

$ L_1 < L_2 < L_3 < L_4 < L_5, $

we want $L_3 = \text{M}$. The English alphabet has 26 letters, and M is the $13^\text{th}$.


Step 1: Letters before M

The letters before M are $\{A, B, C, \ldots, L\}$.

There are 12 letters here ($A$ through $L$).

We need to pick 2 of these 12 letters to occupy $L_1$ and $L_2$.

The number of ways to choose 2 letters out of 12 is ${ }^{12} \mathrm{C}_2$.

Step 2: Letters after M

The letters after M are $\{N, O, P, \ldots, Z\}$.

There are 13 letters here ($N$ through $Z$).

We need to pick 2 of these 13 letters to occupy $L_4$ and $L_5$.

The number of ways to choose 2 letters out of 13 is ${ }^{13} \mathrm{C}_2$.

Step 3: Multiply the choices

Since these choices are independent (picking the two letters before M and two letters after M), the total number of ways is:

$ { }^{12} \mathrm{C}_2 \;\times\;{ }^{13} \mathrm{C}_2 $

Calculate each combination:

$ { }^{12} \mathrm{C}_2 = \frac{12 \times 11}{2} = 66, \quad { }^{13} \mathrm{C}_2 = \frac{13 \times 12}{2} = 78. $

So,

$ { }^{12} \mathrm{C}_2 \times { }^{13} \mathrm{C}_2 = 66 \times 78 = 5148. $


Answer: 5148 (Option B)

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