Question

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

• A. 3
• B. 5
• C. 7
• D. 0

Answer 1

The correct option is A 3

$\left|\begin{array}{ccc}1& 1& 1\\ {}^m{C}_1& {}^{m+3}{C}_1& {}^{m+6}{C}_1\\ {}^m{C}_2& {}^{m+3}{C}_2& {}^{m+6}{C}_2\end{array}\right|=2^{\alpha }{3}^{\beta }{5}^{\gamma }$
$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ m& m+3& m+6\\ \frac{m(m-1)}{2}& \frac{(m+3)(m+2)}{2}& \frac{(m+6)(m+5)}{2}\end{array}\right|=2^{\alpha }{3}^{\beta }{5}^{\gamma }$
Consider,$\left|\begin{array}{ccc}1& 1& 1\\ m& m+3& m+6\\ \frac{m(m-1)}{2}& \frac{(m+3)(m+2)}{2}& \frac{(m+6)(m+5)}{2}\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\left|\begin{array}{ccc}0& 0& 1\\ -3& -3& m+6\\ -3(m+1)& -3(m+4)& \frac{(m+6)(m+5)}{2}\end{array}\right|$
$=3^2\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& m+6\\ (m+1)& (m+4)& \frac{(m+6)(m+5)}{2}\end{array}\right|$
$=3^2[m+4-m-1]=3^3$
So, $2^0{3}^3{5}^0=2^{\alpha }{3}^{\beta }{5}^{\gamma }$
On comparing,
$\alpha =\gamma =0,\beta =3$
So,$\alpha +\beta +\gamma =3$