Q169 / 270 · NDA 2024 (Phase 1) · Mathematics

For what value of $n$ is determinant $\left|\begin{array}{ccc} \mathrm{C}(9,4) & \mathrm{C}(9,3) & \mathrm{C}(10, n-2) \\ \mathrm{C}(11,6) & \mathrm{C}(11,5) & \mathrm{C}(12, n) \\ \mathrm{C}(m, 7) & \mathrm{C}(m, 6) & \mathrm{C}(m+1, n+1) \end{array}\right|=0$ for every $m \gt n$ ?

  • A
    4
  • B
    5
  • Correct answer
    C
    6
  • D
    7

Solution

$\left|\begin{array}{ccc}\mathrm{C}(9,4) & \mathrm{C}(9,3) & \mathrm{C}(10, n-2) \\ \mathrm{C}(11,6) & \mathrm{C}(11,5) & \mathrm{C}(12, n) \\ \mathrm{C}(m, 7) & \mathrm{C}(m, 6) & \mathrm{C}(m+1,7)-\mathrm{C}(m+1, n+1)\end{array}\right|=0$ $\begin{aligned} & \text { Applying } \mathrm{C}_3 \rightarrow \mathrm{C}_1+\mathrm{C}_2-\mathrm{C}_3 \\ & \left|\begin{array}{lcc}\mathrm{C}(9,4) & \mathrm{C}(9,3) & \mathrm{C}(10,4)-\mathrm{C}(10, n-2) \\ \mathrm{C}(11,6) & \mathrm{C}(11,5) & \mathrm{C}(12,6)-\mathrm{C}(12, n) \\ \mathrm{C}(m, 7) & \mathrm{C}(m, 6) & \mathrm{C}(m+1,7)-\mathrm{C}(m+1, n+1)\end{array}\right|=0 \\ & \because \text { Value of determinant }=0 \\ & \therefore \mathrm{C}_3=0 \Rightarrow n-2=4 \Rightarrow n=6 .\end{aligned}$

Hence, the correct answer is option: 'C'.

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