JEE Main 2025 — 22 January, Morning Shift (moderate confidence). Previous Year Question.
Problem
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:
Key insight. Because the five chosen letters must always be arranged in alphabetical order, there’s only ever one way to arrange any given set of five letters — which means the entire question is really just “how many ways can the letters be chosen?”, not arranged. Fixing M in the middle then just splits that selection into two independent, smaller selections: letters before M and letters after M.
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Approach
Since alphabetical order is forced, choosing which 5 letters to use automatically determines their order — there’s no separate arrangement step. With M fixed as the middle (third) letter, the two letters before it must come from the 12 letters earlier in the alphabet (A–L), and the two letters after it must come from the 13 letters later in the alphabet (N–Z). These two selections are independent of each other, so the total count is simply the product of two combination counts.
Solution
Step 1 — Split the alphabet around M
The English alphabet has 26 letters. Removing M leaves 25, split into two groups by position relative to M:
- Before M: A through L — that’s 12 letters
- After M: N through Z — that’s 13 letters
Step 2 — Choose 2 letters from before M
Since exactly 2 of the 5 letters must come before M (to leave M in the middle position with 2 letters on each side):
$${}^{12}C_2 = \frac{12\times11}{2} = 66$$
Step 3 — Choose 2 letters from after M
$${}^{13}C_2 = \frac{13\times12}{2} = 78$$
Step 4 — Multiply the two independent choices
Since choosing the “before” letters and the “after” letters are independent decisions, and each combination of choices produces exactly one valid alphabetically-ordered arrangement (with M fixed in the middle):
$$66\times 78 = 5148$$
Answer
$$5148$$
Common mistakes
- Trying to count arrangements separately from selections. Since alphabetical order is mandatory, there’s no additional factor of $5!$ or similar to multiply by — choosing the letters is choosing the arrangement.
- Splitting the alphabet unevenly around M. M is the 13th letter of 26, so there are exactly 12 letters before it and 13 after — miscounting either side (e.g., using 13 and 13) throws off both combination counts.
Practise next
- From all the English alphabets, five letters are chosen and arranged in alphabetical order — find the number of ways the middle letter is ‘K’, adjusting the before/after letter counts accordingly.
Show answer
$4725$. Once five letters are chosen, the alphabetical arrangement is forced — so this is a counting problem about the choice alone.
For $K$ to be the middle letter, exactly two of the chosen letters must come before it and two after. $K$ is the 11th letter, so there are $10$ letters before it and $15$ after.
Hence $\binom{10}{2}\binom{15}{2}=45\times105=4725$.
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