UPSC NDA I 2024 Mathematics Paper

Concept:Use the combinatorial identity $C(n, r) + C(n, r-1) = C(n+1, r)$ to simplify the determinant.Explanation:Start with the given determinant.Perform column operation: replace column $C_1$ with $C_1 + C_2$.Using the identity, the new first column becomes:For row 1: $C(9,4)+C(9,3)=C(10,4)$.For row 2: $C(11,6)+C(11,5)=C(12,6)$.For row 3: $C(m,7)+C(m,6)=C(m+1,7)$.The determinant now has first column $[C(10,4), C(12,6), C(m+1,7)]$.The third column is $[C(10, n-2), C(12, n), C(m+1, n+1)]$.For the determinant to be zero for every $m > n$, the first and third columns must be identical for all such $m$.This requires matching entries:$C(10,4)=C(10,n-2)$, $C(12,6)=C(12,n)$, and $C(m+1,7)=C(m+1,n+1)$.From $C(10,4)=C(10,n-2)$, we have $n-2=4$ or $n-2=6$ (since $C(10,r)=C(10,10-r)$).From $C(12,6)=C(12,n)$, we get $n=6$ (since $12-6=6$).From $C(m+1,7)=C(m+1,n+1)$, equality for general $m$ requires $n+1=7$ i.e. $n=6$.All conditions consistently give $n=6$.Answer:$n = 6$, which corresponds to option C.