Question

If $\left|\begin{array}{ccc}1& 1& 1\\ {}^m{\text{C}}_1& {}^{m+3}{\text{C}}_1& {}^{m+6}{\text{C}}_1\\ {}^m{\text{C}}_2& {}^{m+3}{\text{C}}_2& {}^{m+6}{\text{C}}_2\end{array}\right|=2^{\alpha }{3}^{\beta }{5}^{\gamma }$, then $\alpha +\beta +\gamma$ is equal

• A. $3$
• B. $5$
• C. $7$
• D. none of these

Answer 1

The correct option is A $3$

$\left|\begin{array}{ccc}1& 1& 1\\ {}^m{C}_1& {}^{m+3}{C}_1& {}^{m+6}{C}_1\\ {}^m{C}_2& {}^{m+3}{C}_2& {}^{m+6}{C}_2\end{array}\right|$

Applying $C_3\to C_3-C_2,C_2\to C_2-C_1$

$=\left|\begin{array}{ccc}1& 0& 0\\ {}^m{C}_1& {}^{m+3}{C}_1{-}^m{C}_1& {}^{m+6}{C}_1{-}^m{C}_1\\ {}^m{C}_2& {}^{m+3}{C}_2{-}^m{C}_2& {}^{m+6}{C}_2{-}^m{C}_2\end{array}\right|$

$={(}^{m+3}{C}_1{-}^m{C}_1\left){(}^{m+6}{C}_2{-}^m{C}_2\right)-{(}^{m+3}{C}_2{-}^m{C}_2\left){(}^{m+6}{C}_1{-}^m{C}_1\right)$

$(\frac{m+3}{1}-m)(\frac{(m+6)(m+5)}{2}-\frac{m(m-1)}{2})-(\frac{(m+3)(m+2)}{2}-\frac{m(m-1)}{2})(\frac{m+6}{1}-m)=3^3$

$\therefore \alpha +\beta +\gamma =3$