I have a grid of $9$ squares $\left(3 \times 3\right)$ and bags of coloured counters of $9$ different colours. I can lay one counter per square and I can lay down three counters. Of the three, more than one can be of the same colour. What is the formula I would need to calculate the number of possible permutations?
I can calculate that without colours, there would be 84 outcomes. The next step, involving the colors, is where I am now stuck.
On
My understanding of the problem gives a much smaller number of possible outcomes.
As given in the original post, there are $84 = \binom{9}{3}$ ways to choose the three squares from the grid.
All that remains is to choose a color for the counter in each of these three positions. Evidently the bags contain enough counters to choose any color for any square. So the assignment of colors to the three squares can be done in $729 = 9^3$ ways.
Altogether there would only be $84 \times 729 = 61236$ distinct outcomes. To get the large number suggested by Air Mike, one would need some labelling of the counters in each bag, so that outcomes would be distinguishable based on which counters are chosen in a particular bag. But the problem statement does not indicate how many counters there are in the bags, and the figure of "9 balls" in each bag does not seem supportable. One might be able to produce all the outcomes with only three counters per bag.
On
$$\textbf{:: NEW ANSWER ::}$$
I revisited this question and I believe that none of the solutions suggested are correct. For that, I’m posting a new answer, so they all can be compared.
Suppose we have a squared grid of size $3 \times 3$ (which gives a total of $9$ little squares) and $9$ distinct bags of counters such that each bag contain counters of the same color and different bags correspond to different colours.
Our job is to select $3$ little squares from our grid, select $3$ counters from the bags and place them in the little squares.
For the sake of simplicity (and because there is nothing stated that goes agains this) I will assume that each bag has, at least, $3$ balls. So, it is possible to select $3$ counters from the same bag. Also, I will assume that the counters of the same bag are indistinguishable. (If not, we can adapt this argument in a straightforward manner to make it all work).
Note that from the $9$ little squares available, we must select $3$ of them and this selection is independent from the order. We are only selecting our squares. Hence, that can be done in $9 \choose 3$ distinct ways.
Let’s divide this in three cases.
$$\begin{align*} \text{Case $a)$}: & \quad \text{we select $3$ counters of distinct bags};\\ \text{Case $b)$}: & \quad \text{we select $2$ counters of the same bags};\\ \text{Case $c)$}: & \quad \text{we select $3$ counters of the same bag}. \end{align*}$$
Fro each case, let’s compute the number of distinct ways that there are of selecting $3$ counters.
:: Case $a)$ :: Suppose we select $3$ counters of distinct bags. In practise, we are selecting $3$ distinct colours from a total of $9$ colours. This selection can be done in $9 \choose 3$ distinct ways.
:: Case $b)$ :: Suppose we select $2$ counters of the same bag. Hence, we are selecting $2$ distinct colours from a total of $9$. So, this selection can be done in $9 \choose 2$ different ways.
:: Case $c)$ :: Suppose we select $3$ counters of the same bag. We are only selecting one colour from $9$. Therefore, this selection can be done is $9$ distinct ways.
Our task is not done yet. For each of the previous cases, we have different ways of placing our counters.
:: Case $a)$ :: We have $3$ counters of $3$ distinct colours to be places in $3$ distinct squares. This placement can be done in $3!$ distinct ways.
:: Case $b)$ :: We have $2$ counters of the same colour and a third of a different one. We can place these $3$ counters in $3$ distinct ways. Note that the counters of the same colour can commute and the arrangement will still be the same.
:: Case $c)$ :: We have $3$ counters of the same colour. It doesn’t matter the order that we place them, the arrangement will always be the same. So, we only have one way to do this.
Finally, for each case, we will multiply the number the numbers at points $4$, $5$ and $6$ (this is a consequence of the Rule of Produtct). This will give the total number of distinct ways to place the $3$ counters in $3$ little squares, for each case. Adding those together gives us the result we want.
:: Case $a)$ :: ${9 \choose 3} \times {9 \choose 3} \times 3! = 42336$;
:: Case $b)$ :: ${9 \choose 3} \times {9 \choose 2} \times 3 = 9072$;
:: Case $c)$ :: ${9 \choose 3} \times 9 \times 1 = 756$.
So, there are $52164$ distinct ways of selecting $3$ counters and placing them in $3$ little squares.
$$\textbf{:: OLD ANSWER ::}$$
You have $9$ bags with $9$ different coloured balls. In total, you have $9 \times 9 = 81$ balls. From this $81$ balls, you will choose $3$ balls. So, that gives you $\binom{81}{3} = 85320$ possibilities of choice.
Look at the grid and choose $3$ squares. Note that you have $\binom{9}{3} = 84$ distinct ways to choose the squares from your $3 \times 3$ grid. Together, this two events can occur in $85320 \cdot 84 = 7166880$ different ways.
Now, you just have to place the balls. You have to place $3$ balls in $3$ squares, and the order you placed them is relevant. So you have $3! = 6$ different possibilites of placing the balls.
That gives you a total of $7166880 \cdot 6 = 43001280$ possible and distinct outcomes.