$R1$ has $2$ squares, $R2$ has $4$ squares and $R3$ has $2$ squares
We need to place 6Xs, let's take all possibilities
R1 |
R2 |
R3 |
1 |
4 |
1 |
2 |
3 |
1 |
2 |
2 |
2 |
1 |
3 |
2 |
Total number of ways = $(_{1}^{2}\textrm{C}*_{4}^{4}\textrm{C}*_{1}^{2}\textrm{C})+(_{2}^{2}\textrm{C}*_{3}^{4}\textrm{C}*_{1}^{2}\textrm{C})+(_{2}^{2}\textrm{C}*_{2}^{4}\textrm{C}*_{2}^{2}\textrm{C})+(_{1}^{2}\textrm{C}*_{3}^{4}\textrm{C}*_{2}^{2}\textrm{C})$
$= 2*1*2 +1*4*2 + 1*6*1 + 2*4*1$
$ = 4 + 8 + 6 + 8$
$ = 26$
Alternative way
Selecting $6$ squares out of $8$ squares can be done in $_{6}^{8}\textrm{C} = 28$ ways
This will contain the folwing combination $(R1, R2,R3)\rightarrow ( 2 , 4 , 0 ) $ and $(R1,R2,R3)\rightarrow ( 0 , 4 , 2 ) $, but according to the question we need atlast $1 X$ in each row
$\therefore$ we need to subtract these $2$ combinations from $28$ to get the result.
So, the answer will be $28 - 2 = 26$