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.  
 A: Position would be $(1,1,1,1,2)$ or $(1,1,2,1,1)$ or $(2,1,1,1,1)$ for five rows respectively.
Positions to which can beselected in $\binom 32$ for the row with $2$ elements and $3\times3$ for other two.
Now treating each symbol as a unique entity, arrangement can be done in $6!$ ways. Now to cancel  multiples we divide by $2!2!$ one each for + and #. So ways are:
$$3\times\left[\binom 32\times3\times3\right]\times \frac14\times6!=14580\text{ ways}$$
A: 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
A: First, this problem is not as hard as it looks.  It appears reasonable to assume that the symbols are used up as they are placed.  For now, we can think of each symbol as being unique.  Since the symbols are placed "so that no row remains empty", and since there are 6 symbols, we have:
    6 ways to put a symbol into row 1
    5 ways to put a symbol into row 2
    ...
    2 ways to put a symbol into row 5

Note that only 3 of the rows has room for a final symbol.  Thus there are:
    3 ways to choose one of 3 rows to place the final symbol

We have over-counted due to the fact that there are duplicate #, # and +, +.  So we must divide by 2*2.
Finally the number of ways is given by:
(6*5*4*3*2 * 3) / (2*2) = 540

So the answer is 540 ways.
