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#### Please solve RD Sharma class 12 chapter 5 Determinants exercise multiple choise question 33 maths textbook solution

Correct option (c)

Hint:

Using combination properties

$i.e. \: \: nC_{1}=n\: \: and\: \: nC_{2}=\frac{n(n-1)}{2}\: \: and \: \: so\: \: on.$

Given:

We have

$\left | 1\: 1\: 1\: nC_{1}n+ 2C_{1}n+4C_{1}\: nC_{2}n+2C_{2}n+4C_{2} \right |$

We have to find the value of given determinant.

Solution:

We have,

$\left|\begin{array}{ccc} 1 & 1 & 1 \\ n C_{1} & n+2 C_{1} & n+4 C_{1} \\ n C_{2} & n+2 C_{2} & n+4 C_{2} \end{array}\right|$

We know that

\begin{aligned} &n C_{1}=n, n C_{2}=\frac{n(n-1)}{2} n+2 C_{2}=\frac{(n+2)(n+1)}{2} \quad \text { and } \quad n+4 C_{2}=\frac{(n+4)(n+3)}{2}\\ &\left|\begin{array}{ccc} 1 & 1 & 1 \\ n C_{1} & n+2 C_{1} & n+4 C_{1} \\ n C_{2} & n+2 C_{2} & n+4 C_{2} \end{array}\right|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ n & n+2 & n+4 \\ \frac{n(n-1)}{2} & \frac{(n+2)(n+1)}{2} & \frac{(n+4)(n+3)}{2} \end{array}\right| \end{aligned}

Applying C2 → C2 - C1; C3 → C3 - C1

$=\left|1\: 0\: 0 \: n\: 2\: 4 \frac{n(n-1)}{2} 2\: n+1\: 4 n+6\right|$

Expanding along R1

$=2(4n+6)-4(2n+1)$

$=8n+12-8n-4$

$=8$

$Hence,\: \: \left|\begin{array}{ccc} 1 & 1 & 1 \\ n C_{1} & n+2 C_{1} & n+4 C_{1} \\ n C_{2} & n+2 C_{2} & n+4 C_{2} \end{array}\right|=8$