Question

If total $40\lambda ,\lambda \in N$ matrices of different orders can be formed by using all the roots (counting multiplicity) of equation $(x-1{)}^3(x-2{)}^3(x-3{)}^2=0$ exactly once as entries, then the value of $\lambda$ is

Answer 1

Roots of equation $(x-1{)}^3(x-2{)}^3(x-3{)}^2=0$ are $1,1,1,2,2,2,3$ and $3$
So, total elements $=8$
Possible orders are $1\times 8,8\times 1,2\times 4$ and $4\times 2$
Total ways of arranging all roots $=\frac{8!}{3!\cdot 3!\cdot 2!}=560$
So, total different matrices $=4\times 560$
$\Rightarrow 40\lambda =4\times 560$
$\therefore \lambda =56$