Placing symbols so that no row remains empty. 
Q.1.The symbols +, + , #, # , *, $ ($6$ in total) are to be placed in the squares of the given figure. Find the number of ways of placing the symbols so that no row (there are $5$ rows) remains empty.
I know combinatorics at undergrad level. This problems are from advanced challenging section.  

Not sure but I would say : $3\times3\times3\times3 \times 6!$
If the symbols were the same: you are forced to place $2$ symbols on the one-case rows
Then you have to place your last $4$ symbols. 
$3$ rows  with $3$ cases left so it means two rows with $1$ symbol and $1$ row  with two symbols. 
The number of possibilities to place 1 symbol in one of $3$ cases is the same than 2 symbols in two of 3 cases $(\frac{3!}{1!(3-1)!}=\frac{3!}{2!(3-2)!}  = 3)$
So basically for each one of these $3$ rows, you have $3$ possibilities :$3\times3\times3 $
Moreover to select the row that will receive $2$ symbols, you have $3$ possibilities :$3\times3\times3\times3$
So $3\times3\times3\times3$ is the number of possibilities to select cases. $6!$ for the number of permutations
But i might be wrong
