Question

If total $180n,n\in N$ different matrices can be formed by using all the roots of equation $(x-1{)}^2(x-2{)}^3(x-3{)}^4=0.$ Then the value of $n$ is

Answer 1

The roots of equation $(x-1{)}^2(x-2{)}^3(x-3{)}^4=0$ are $1,1,2,2,2,3,3,3,3.$
So, total elements $=9$
Possible orders $=1\times 9,9\times 1,3\times 3$
Total ways of arranging all roots $=\frac{9!}{2!\cdot 3!\cdot 4!}$
$=1260$
So, total different matrices $=3\times 1260=3780$
$\Rightarrow 180n=3780$
$\therefore n=21$