\[a = 1 - \frac{\binom{3}{5}}{\binom{12}{5}}\]
\[b = 3 \cdot \frac{\binom{9}{4}}{\binom{12}{5}}\]
\[c = 3 \cdot \frac{\binom{9}{3}}{\binom{12}{5}}\]
\[d = 1 \cdot \frac{\binom{9}{2}}{\binom{12}{5}}\]
\[u = 0 \cdot a + 1 \cdot b + 2 \cdot c + 3 \cdot d = 1.25\]
\[\sigma^2 = 0 \cdot a + 1 \cdot b + 4 \cdot c + 9 \cdot d - u^2\]
\[\sigma^2 = \frac{105}{176}\]
\[\text{Ans.} \quad 176 - 105 = 71\]
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: