Number of ways to arrange 4 identical decks of cards Let's say we have 4 identical decks of cards, and each one has 10 cards (therefore we have 4 units of the first card, 4 units of the second card and so on). In how many ways can we arrange these 40 cards?
I think the result is the following:
$$\begin{pmatrix} 40 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 36 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 32 \\ 4 \end{pmatrix} \cdots \begin{pmatrix} 4 \\ 4 \end{pmatrix} = \begin{pmatrix} 40 \\ 4,4,4,\ldots,4 \end{pmatrix}= \frac{40!}{(4!)^{10}}$$
Is this correct?
 A: Your solution is correct (and the method you chose to arrive at it is also very logical). Here is an alternative method:
Say you need to arrange $p$ objects, and $q$ of those objects are identical, while the other $p-q$ objects are all different from each other. The number of combinations is
$$
\frac{p!}{q!}
$$
To make things more concrete and clear, here is an example. You need to put $5$ cardboard shapes on a shelf. Consider the case where all $5$ of the shapes are different from each other:

Here, you have $5$ options for the first spot on the shelf; then, when you have placed a shape on the shelf, you have $4$ options left for the second spot on the shelf. This is the rationale behind using factorials: they allow us to concisely denote $n \times (n-1)  \times (n-2) \times \dots \times 3 \times 2 \times 1$. Now that we have the more basic stuff out of the way, let's consider the case where $2$ or more of the objects are identical: say, $3$ triangles, $1$ square, and $1$ pentagon. Without any matching shapes, the number of possible arrangements would be equal to $5!=120$. However, because you have $3$ identical triangles, we have to divide by $3!$ to get $20$ possible arrangements. To explain why, imagine we listed all $120$ possible combinations (ignoring any duplicates for now). Let's denote the items in our list as $\{t_1, t_2, t_3, s, p\}$. $t_1$ and $t_2$ actually look the same, but let's imagine we kept track of where they are. If we listed all the possible arrangements, then the list might look a bit like this:
$$
\{t_1, t_2, t_3, s, p\} \\
\{t_1, t_3, t_2, s, p\} \\
\{t_2, t_1, t_3, s, p\} \\
\{t_2, t_3, t_1, s, p\} \\
\{t_3, t_1, t_2, s, p\} \\
\{t_3, t_2, t_1, s, p\} \\
\dots
$$
As you can see, a single combination has $6$ different 'forms'. To count the number of duplicates, once again we can use factorials. There are $3$ choices for which triangle we use first, then $2$ choices, then $1$ choice. Therefore, we divide by $3!$, which is indeed equal to $6$.
Therefore, the '$q!$' in the $\frac{p!}{q!}$ removes duplicates. Things get a little bit more complicated when you consider something like $4$ identical squares, $5$ identical pentagons, and $7$ identical heptagons, but the same rule applies. We divide by $(4! \times 5! \times 7!)$ because for each shape, there are duplicates that we need to remove. Try listing out the combinations in the way we have done above if it still doesn't make sense.
Incidentally, this is how we arrive at the $\binom{n}{r}$ formula. Let's say that there are $n$ shapes that need to be arranged: $r$ circles and $n-r$ squares .The formula is
$$
\frac{n!}{r!(n-r)!}
$$
Finally, we get to your question. Here, there are $10$ groups of $4$ objects. In each group, all of the objects are identical. Therefore, your answer can be found more directly using
$$
\frac{40!}{4!\times4!\times4!\times4!\dots\times4!}=\frac{40!}{(4!)^{10}}
$$
If you have any questions about why these formulae work, then feel free to ask.
