P and C – Word Problem – 22 January 2025 (Shift 1)

Question ID: #128
JEE Main22 January Shift 1, 2025Algebra

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:

  • (1) 14950
  • (2) 6084
  • (3) 4356
  • (4) 5148

Solution:

We need to choose 5 letters such that when arranged alphabetically, the 3rd (middle) letter is ‘M’.

$$ \underline{L_1} \quad \underline{L_2} \quad \mathbf{M} \quad \underline{L_4} \quad \underline{L_5} $$

For the arrangement to be alphabetical:

  • $L_1, L_2$ must be chosen from letters strictly before ‘M’ (A to L). Total 12 letters.
  • $L_4, L_5$ must be chosen from letters strictly after ‘M’ (N to Z). Total 13 letters.

Since the final arrangement is fixed (alphabetical order), we only need to select the letters.

Number of ways to select 2 letters from first 12 = $^{12}C_2$.

Number of ways to select 2 letters from last 13 = $^{13}C_2$.

Total ways:

$$ ^{12}C_2 \times ^{13}C_2 = \frac{12 \times 11}{2} \times \frac{13 \times 12}{2} $$

$$ = 66 \times 78 = 5148 $$

Ans. (4)

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