Poker and Combinatorics (Don't Mix): How to solve this problem?

Does anyone know the solution to this problem? I'm not sure if it's asking to find the cardinality first, and whether I should use the additive counting principle or the multiplicative one? Would help me a lot, thanks!

Given is a standard $52$-card poker game. A hand is a non-ordered selection of five Cards from the game. Consider the following question and the answer to it:

Question: How many hands are there that have all four colors?

Answer: We choose a card from each color, so there are $(13^4)$ possibilities for this. $48$ possibilities remain when we choose the fifth card. So there are a total of $(13^4) \times 48$ hands that contain all four suits.

Is the solution correct? If not, what is wrong with the answer and what is the correct solution?

There will be double-counting in your result. In fact, every choice is counted twice: suppose two of your cards are chosen to be e.g. diamonds. That choice is counted twice: once with the first diamond card chosen as you did your first choice of four cards (and the second chosen as the "fifth" card), and once the other way round.

Thus, the total number is half of what you've got, i.e.

$$\frac{1}{2}\left(13^4\times 48\right)=685,464$$

You can see this also the following way: first pick a suit ($4$ ways). Pick two cards from that suit ($13\choose 2$ ways). Pick one card from each of the three remaining suits ($13^3$ ways). The total is:

$$4{13\choose 2}\times 13^3=685,464$$

which is the same as we got before.

When you first choose $4$ cards of different colors and then multiply by $48$, it brings in duplicates. So one of the correct approaches would be -

$ \displaystyle 4 \cdot 13^3 \cdot {13 \choose 2}$

Explanation: Choose one of the four colors that you will have two cards of, choose two cards of that color and one card each of rest three colors.