Mathematics - Matrices Question with Solution | TestHub

Given Matrix $3\times 3=\left[\begin{array}{ccc}-& -& -\\ -& -& -\\ -& -& -\end{array}\right]$

Here we can see we have to find nine elements whose sum is $5$

Case 1: When five 1's are there & four O's are there 

Now by division & distribution Method

We get $\Rightarrow \frac{9!}{5!4!}=126 ...\left(\mathrm{i}\right)$

Case 2: when six 1's are there & one $\left\{-1\right\}$ & two O's are there.

Again by division & distribution Method

we get $\Rightarrow \frac{9!}{6!2!1!}=252 ...\left(\mathrm{ii}\right)$

Case 3: When seven 1's are there are two $\left\{-1\right\}$ are there,

By division & distribution we get

$\Rightarrow \frac{9!}{7!2!}=36 ...\left(\mathrm{iii}\right)$

Now adding equation $\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)+\left(\mathrm{iii}\right)$

We get $126+252+36=414$

So total $414$ ways will be there.