<p>\(\begin{vmatrix} ^xC_r & ^xC_{r+1} & ^xC_{r+2} \\ ^yC_r & ^yC_{r+1} & ^yC_{r+2} \\ ^zC_r & ^zC_{r+1} & ^zC_{r+2} \end{vmatrix}\) is equal to</p>
<p>\(\begin{vmatrix} ^xC_r & ^{x+1}C_{r+1} & ^{x+1}C_{r+2} \\ ^yC_r & ^{y+1}C_{r+1} & ^{y+1}C_{r+2} \\ ^zC_r & ^{z+1}C_{r+1} & ^{z+1}C_{r+2} \end{vmatrix}\)</p>
<p>\(\begin{vmatrix} ^xC_r & ^{x+1}C_{r+1} & ^{x+2}C_{r+2} \\ ^yC_r & ^{y+1}C_{r+1} & ^{y+2}C_{r+2} \\ ^zC_r & ^{z+1}C_{r+1} & ^{z+2}C_{r+2} \end{vmatrix}\)</p>
<p>\(\begin{vmatrix} ^xC_r & ^xC_{r+1} & ^{x+1}C_{r+2} \\ ^yC_r & ^yC_{r+1} & ^{y+1}C_{r+2} \\ ^zC_r & ^zC_{r+1} & ^{z+1}C_{r+2} \end{vmatrix}\)</p>
<p>\(\begin{vmatrix} ^xC_r & ^{x+1}C_{r+1} & ^xC_{r+2} \\ ^yC_r & ^{y+1}C_{r+1} & ^yC_{r+2} \\ ^zC_r & ^{z+1}C_{r+1} & ^zC_{r+2} \end{vmatrix}\)</p>
Step-by-Step Solution
Key Concept: Use the property that C(n,r+1)/C(n,r) = (n-r)/(r+1) to establish linear relationships between columns, then apply column operations to reduce the determinant to a product form involving differences of binomial coefficients.
<p><strong>Step 1:</strong> Recognize the structure—each column contains binomial coefficients with the same r value across different n values (x, y, z).</p><p><strong>Step 2:</strong> Use column operation C₂ → C₂ - C₁ and C₃ → C₃ - C₂. The binomial coefficient identity gives:<br>C(n,r+1) - C(n,r) = C(n,r)·[(n-r)/(r+1) - 1]</p><p><strong>Step 3:</strong> After column operations, each element in the new columns becomes expressions involving differences. The determinant becomes proportional to:<br>∏[C(n,r) · (n-r-r-1)] for different n values.</p><p><strong>Step 4:</strong> The determinant factors as:<br>Δ = C(x,r)·C(y,r)·C(z,r)·(y-x)(z-x)(z-y) / [(r+1)!]<br>Or in standard form: <strong>Δ = C(x,r)·C(y,r)·C(z,r)·(y-x)(z-y)(z-x)</strong></p><p><strong>Step 5:</strong> This can also be expressed as:<br>Δ = <sup>x</sup>C<sub>r</sub>·<sup>y</sup>C<sub>r</sub>·<sup>z</sup>C<sub>r</sub>·(x-y)(y-z)(z-x) / [r!(r+1)!]<br>Or simply: <strong>C(x,r)·C(y,r)·C(z,r)·(y-x)(z-y)(z-x)</strong></p><p>∴ Answer: C</p>
Correct Answer: C