Combinations of two sets of five elements, or five sets of two elements? That’s advance maths for my highschool level, sorry guys for such lame question
Let’s say I have a set of five cards (ABCDE), I know , the number of possible combinations is $5\cdot 4\cdot 3\cdot 2\cdot 1 = 120$
Now I have two sets (booths equal cards), one red, one blue, and I drop first cards from one set (red), then from the second (blue)

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*then it will be $5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 = 14400$ $(120^2)$
But how to calculate if I mix both set, and I place them randomly? (without taking care of the color)

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*First row: 5 options

*Second row: 5 options also (with different distribution of possibilities)

*Third row, depends:

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*If first two cards are the same: then 4 option

*if first two cards are different : then 5 option



*Fourth row, depends a lot...

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*and at this point I start to get lost



And suspicions, tell me that result will be different if I take five decks of two cards each (always the same)
Not sure how should I search for the solution, therefore I ask here
 A: 
How many ways can two decks of five cards, each labeled from A to E, be arranged in a row if two arrangements are distinguished only by the positions of the letters?

Choose two of the ten positions for the As, two of the remaining eight positions for the Bs, two of the remaining six positions for the Cs, two of the remaining four positions for the Ds, and fill the last two positions with Es, which can be done in
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{10!}{2!8!} \cdot \frac{8!}{2!6!} \cdot \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{10!}{2!2!2!2!2!}$$
The factors of $2!$ in the denominator represent the number of ways identical letters can be permuted among themselves within an arrangement without creating an arrangement distinguishable from the given arrangement.

How many ways can five decks of two cards, each labeled from A to B, be arranged in a row if two arrangements are distinguished only by the positions of the letters?

Such an arrangement is completely determined by choosing five of the ten positions for the As since the remaining five positions must be filled with Bs.  Hence, there are
$$\binom{10}{5} = \frac{10!}{5!5!}$$
such arrangements.  Notice that the factors in the denominator are the number of ways we can permute the As among themselves and the Bs within themselves within an arrangement without creating an arrangement distinguishable from the given arrangement.
