There are 26 letters in the English alphabet (A to Z).
We are choosing 5 letters such that when they are arranged in alphabetical order, the middle letter is ‘M’, i.e., the third letter.
Once we fix 'M' as the middle letter, we need to choose:
2 letters from the letters before ‘M’ (i.e., A to L → total 12 letters)
2 letters from the letters after ‘M’ (i.e., N to Z → total 13 letters)
The number of ways to choose 2 letters from 12 before M is:
\( \binom{12}{2} = 66 \)
The number of ways to choose 2 letters from 13 after M is:
\( \binom{13}{2} = 78 \)
Since the letters must be in alphabetical order, we don't need to arrange them — each unique set leads to only one valid arrangement. So:
\[ \text{Total ways} = \binom{12}{2} \times \binom{13}{2} = 66 \times 78 = 5148 \]
Option 4: 5148
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to:
Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \text{Im}(\alpha) < \text{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \[ P_3 = -5\sqrt{7}, \quad P_4 = -3\sqrt{7}, \quad P_5 = 11\sqrt{7}, \quad P_6 = 45\sqrt{7}, \] then \( |\alpha^4 + \beta^4| \) is equal to: