The value of the determinant mmm nC1 n+1C1 n+2C1 nC2 n+1C2 n+2C2 is equal to

Δ=m111 nC1 n+1C1 n+2C1 nC2 n+1C2 n+2C2=m111nn+1n+2n(n−1)2(n+1)n2(n+2)(n+1)2C2↔C2−C1C3↔C3−C2=m100n11n(n−1)2nn+1=m.

The value of the determinant mmm nC1 n+1C1 n+2C1 nC2 n+1C2 n+2C2 is equal to

Δ=m111 nC1 n+1C1 n+2C1 nC2 n+1C2 n+2C2=m111nn+1n+2n(n−1)2(n+1)n2(n+2)(n+1)2C2↔C2−C1C3↔C3−C2=m100n11n(n−1)2nn+1=m.

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The value of the determinant $|\begin{array}{ccc} m & m & m \\ ^{n} C_{1} & ^{n + 1} C_{1} & ^{n + 2} C_{1} \\ ^{n} C_{2} & ^{n + 1} C_{2} & ^{n + 2} C_{2} \end{array}|$ is equal to

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$m$

$m n$

$n (n - 1)$

answer is A.

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Detailed Solution

$Δ = m |\begin{array}{ccc} 1 & 1 & 1 \\ ^{n} C_{1} & ^{n + 1} C_{1} & ^{n + 2} C_{1} \\ ^{n} C_{2} & ^{n + 1} C_{2} & ^{n + 2} C_{2} \end{array}|$

$\begin{array}{l} = m |\begin{array}{ccc} 1 & 1 & 1 \\ n & n + 1 & n + 2 \\ \frac{n (n - 1)}{2} & \frac{(n + 1) n}{2} & \frac{(n + 2) (n + 1)}{2} \end{array}| \begin{array}{c} C_{2} \leftrightarrow C_{2} - C_{1} \\ C_{3} \leftrightarrow C_{3} - C_{2} \end{array} \\ = m |\begin{array}{ccc} 1 & 0 & 0 \\ n & 1 & 1 \\ \frac{n (n - 1)}{2} & n & n + 1 \end{array}| = m . \end{array}$