Number of the ways to place six X's in a specific arrangement of squares.

The question says: Six X's have to be placed in the squares of figure below in such a way that each row contains atleast one X. In how many ways can this be done? The figure:

My approach: Let's first ensure that each row has atleast one X. So

Number of ways to place one X in top row: ${2 \choose 1} = 2$

Number of ways to place one X in middle row: ${4 \choose 1} = 4$

Number of ways to place one X in bottom row: ${2 \choose 1} = 2$

Number of ways I can redistribute remaining 3 X's in 5 remaining boxes ${5 \choose 3} = 10$

Thus total number of ways are: $2 \cdot4\cdot2\cdot10 = 160 $

But this answer has some flaw which I am unable to get. Kindly tell what exact mistake is there in my approach?

Total ways to select 6 places out of $8$ places $\binom{8}{6}=\binom{8}{2}=28$.

Number of ways to select 6 places so that first row is empty: $1$ (since we have only 6 places remaining.)

Number of ways to select 6 places so that second row is empty: $0$ (since we have to select 6 from remaining 4 places which is not possible.)

Number of ways to select 6 places so that third row is empty: $1$

Hence the answer is $28-(1+0+1)=26$