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UPSC NDA I 2024 Mathematics Paper

Question : 19 of 120

Marks: +1, -0

n

\begin{vmatrix} C(9,4) & C(9,3) & C(10, n-2) \\ C(11,6) & C(11,5) & C(12, n) \\ C(m, 7) & C(m, 6) & C(m+1, n+1) \end{vmatrix}=0

m>n

Solution:

Concept:

Combination formula:

C(n, r)+C(n, r-1)=C(n+1, r)

Explanation:

\begin{vmatrix} C(9,4) & C(9,3) & C(10,n-2) \\ C(11,6) & C(11,5) & C(12,n) \\ C(m,7) & C(m,6) & C(m+1,n+1) \end{vmatrix}=0

Using

C_1 = C_1 + C_2

and using the formula

\begin{vmatrix} C(10,4) & C(9,3) & C(10,n-2) \\ C(12,6) & C(11,5) & C(12,n) \\ C(m+1,7) & C(m,6) & C(m+1,n+1) \end{vmatrix}=0

This determinant will be zero if 1st and 3rd column are same, which is possible only when

n-2=4 \; \text{i.e.,} \; n=6

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