QJEEM25Ma30ec: JEE Main 2025 Mathematics - Class 11 JEEM25Ma30ec Solution: From all the English alphabets, five letters are chosen and are …

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)