$$\begin{vmatrix} ^5C_1 & ^5C_2 & ^5C_3 \\ ^6C_1 & ^4C_2 & ^4C_3 \\ ^3C_1 & ^3C_2 & ^3C_3 \end{vmatrix} \begin{vmatrix} ^5C_1 & ^6C_2 & ^7C_3 \\ ^4C_1 & ^5C_2 & ^6C_3 \\ ^1C_1 & ^4C_2 & ^5C_3 \end{vmatrix} = $$ _____.
Step-by-Step Solution
Key Concept: Recognize that the first determinant has linearly dependent rows due to the binomial coefficient identity ^nC_{r-1} + ^nC_r = ^{n+1}C_r, making it zero. Once one determinant in the product is zero, the entire product equals zero regardless of the second determinant's value.
The binomial coefficient relation $^nC_{r-1} + ^nC_r = ^{n+1}C_r$ is a fundamental identity in combinatorics that follows from $\binom{n}{r-1} + \binom{n}{r} = \frac{n!}{(r-1)!(n-r+1)!} + \frac{n!}{r!(n-r)!} = \frac{n!(r + n - r + 1)}{r!(n-r+1)!} = \binom{n+1}{r}$.
Correct Answer: 0