Total number of possible arrangements without keeping any row devoid of elements

Question

I am given with 6 symbols- +, +, x, x, ☆, ●. How many possible arrangements of these symbols into the below figure is possible if no row should be empty?

I have tried to proceed with this in several ways like finding total number of possible arrangements and then subtracting by those not in the same rows. But it won't work.

Also, it is pretty clear that the boxes in second and fourth row have to be filled at all events. So, we are left to place one element in 6 boxes. I choose and permute but at the end the answer still doesn't come right.

Answer 1

$\underline{Method\;\; avoiding\;\; casework:}$

Taking the symbols to be $AABBCD$ and straightening the $\Bbb{E}$ pattern for simplicity,
permute the symbols and put them in $5$ boxes, e.g. one permutation could read $\boxed{AB-}\boxed{D}\boxed{B--}\boxed{C}\boxed{-A-}$

There are $\frac{6!}{2!2!} = 180$ permutations of symbols,
$3\binom32 = 9$ ways of choosing box for two symbols, and placing them,
and $3\cdot3 = 9$ ways for placing a symbol in the other two "triple" boxes

Putting it all together, we get a total of $180\cdot81$ ways

Answer 2

To make matters clearer, I will refer to the symbols with colors as well: $\color{red}{+,+,\times,\times},\color{blue}{\star,\bullet}$

To begin, make the following observations

• As there are five rows and six symbols, one row must have two symbols while the others each have one.

• The row which has two symbols must be one of those with three available spaces.

Let us break into cases based on what the two symbols appearing in the row with two symbols actually are. The cases are:

• Both are red and the same symbol ($\color{red}{++}$ or $\color{red}{\times \times}$)
• Both are red and different symbols ($\color{red}{+\times}$)
• One is red and one is blue (e.g. but not limited to: $\color{red}{+}\color{blue}{\star}$, $\color{red}{\times}\color{blue}{\star}$...)
• Both are blue ($\color{blue}{\star\bullet}$)

Calculate the number of possibilities for each case.

E.g. in the first listed case:

• Pick which symbol is repeated in the row: $2$ choices
• Pick which row the repeated symbol goes into: $3$ choices
• Pick which space in the row with repeated symbol is left unoccupied: $3$ choices (equivalently worded, pick how to arrange the symbols and the empty space in the repeated row: $\frac{3!}{2!1!}$)
• Pick how to distribute remaining symbols among the remaining four rows: $\frac{4!}{2!1!1!} = 12$ choices (equivalently worded, pick which row gets the star, then which gets the bullet, the remaining red symbols will take whichever is left)
• For those rows with three spaces remaining, pick which of the spaces is occupied by the symbol: $3\cdot 3 = 9$ choices

There are then $2\cdot 3\cdot 3\cdot 12\cdot 9$ arrangements in the first case. Approach similarly for the other cases.