How many possible deals in straight poker.

Question

Each player is dealt 5 cards. 5 players how many deals are possible? I know that for one player there is 2598960 possible outcomes i.e. 52 C 5, I need to know how I can do this for the next four players and why if possible

Answer 1

Hint: having chosen five cards for the first player, how many cards are left? Now choose the second player's hand, multiply that by the number of first player hands and you have the number of first and second player hands. Continue.

Answer 2

The five hands are set as soon as the deck is shuffled/cut for the last time. So, as commented by @Pocho la Pantera:

How many ways can the entire deck be arranged? $\text{ }52!$

Does it matter how the five cards destined for Player $\text{#}1$ are arranged? No, so divide by the number of such arrangements, $\text{ }5!$.

But the same is true for each of the five players, so divide by this factor a total of five times $\text{ }(5!)^5$

Does it matter how the $27$ cards that don't even get dealt are arranged? No, so divide by $\text{ }27!$

Thus the answer is $$\frac{52!}{(5!)^527!}$$