Six x’s has to be placed in the squares in the adjacent figure, such that each row contains at least one x. This can be done in 
Question is - 

Six x’s has to be placed in the squares in the adjacent figure, such
  that each row contains at least one x. This can be done in how many ways?

Due to the limit on the number of x's, I find this question very diffiult to solve. 
I tried representing the figure as $$_,_|_,_,_,_|_,_$$
And 3 x's are already use for each row and we are left with 3 more x's
but I'm unable to proceed further. How to tackle such problems?
 A: See there are $8$ places to be filled but even though if we fill only the rows then also $2$ places would aleays remain. So to arrange 6 letters among 8 places its simply ${8\choose 6}-2=26$ .$-2$ as two of the top or bottom  places would never be filled howsoever we place $x$ hope its clear.
A: Let's count cases based on how many cells of the middle row are occupied


*

*All $4$ cells. Then you have $2$ remaining $x$'s and you should put $1$ in the first row ($2$ options) and $1$ in the third row (another $2$ options). So $$2\times 2=4$$ ways.

*$3$ cells. You can occupy $3$ cells in $\dbinom{4}{3}=4$ ways. This leaves you with $3$ more $x$'s. Choose now a row from rows $1$ and $3$ ($2$ ways to do that), put $2$ of the $x$'s in that row and put the last $x$ in the remaining row ($2$ cells to choose from). So $$4\times 2 \times 2=16$$ ways.

*$2$ cells. You can occupy $2$ cells in $\dbinom{4}{2}=6$ ways. This leaves you with $4$ more $x$'s and no other choice but to place them in the remaining cells. So $$6\times 1=6$$ ways.


If you occupy less than $2$ cells in the middle row, then there is not enough space for the remaining $x$'s, so that was all. Summing up $$4+16+6=26$$ ways.
A: If $n_i$ denotes the number of x's in row $i$ diminished by $1$ then $n_1+n_2+n_3=3$ where $n_1,n_3\in\{0,1\}$ and $n_2\in\{0,1,2,3\}$. Discern the following cases:


*

*$0+3+0$ gives $\binom21\binom44\binom21=4$ possibilities.

*$1+2+0$ gives $\binom22\binom43\binom21=8$ possibilities.

*$0+2+1$ gives $\binom21\binom43\binom20=8$ possibilities.

*$1+1+1$ gives $\binom22\binom42\binom22=6$ possibilities.
$4+8+8+6 =26$ possibilities.

A: Only the top or bottom row can be left blank in $1$ way each,
thus ways with at least one  $x$ in each row = $\binom86 -2 = 26$
A: Here is an approach via generating functions.  Consider the more general problem of placing $r$ x's in the figure.  Let $a_r$ be the number of ways this can be done, subject to the constraint of having at least one x in each row.  Then the generating function for $\{a_r\}$ is
$$f(x) = [(1+x)^2-1]^2 \cdot [(1+x)^4-1] = x^8+8 x^7+26 x^6+44 x^5+40 x^4+16 x^3$$
The answer to the original problem is the coefficient of $x^6$, i.e. $26$.
A: If the bottom row only has a left x, then there are 5 more x's to distribute among the top two rows. We are then essentially choosing which box does not have an x (as there are 6 boxes), and any such choice is valid, so there are 6 choices. The same count holds when the bottom row only has a right x. Now we consider when both bottom boxes have an x. Say then the top row only has a left x. Then there are 3 remaining x's to go to the middle row, so there are 4 many options. The same for the top row only having a right x. The last situation is when the top row is filled, and there are 2 x's to distribute in the middle row. There are 6, 4 choose 2, many options.
6+6+4+4+6 = 26.
