Assume there are 10 different playing cards to be distributed among 2 players (a player may get no card also). Then the total number of ways it can be done is : $ 2^{10} = 1024$. In general case, $r^n$
However, assume the cards have numbers written on it and their order also matters. Hence, the distribution
Player A : 2
Player B : 1 4 3
is different from
Player A : 2
Player B : 3 1 4
Now, if the order of cards matters, what is the total possible number of ways?
If we want the first player to have $i$ cards (and the second player $10-i$), the number of ways is $\binom{10}{i}$. Now, there are $i!$ ways to permute the cards of the first player, and $(10-i)!$ to permute the cards of the second player. So in total the number of ways is $$\sum_{i=0}^{10}\binom{10}{i}i!(10-i)!.$$
This is equal to $11!$. Another way to see this is to think of the problem as permuting the $10$ cards plus one dummy card, where the dummy card serves as a boundary between the set assigned to the first player and that assigned to the second player.
Similarly, for the general case the number of ways is $$\sum_{x_1+\dots+x_r=n}\binom{n}{x_1,x_2,\dots,x_r}x_1!x_2!\dots x_r!.$$ This is equal to $n!\binom{n+r-1}{n}=\frac{(n+r-1)!}{(r-1)!}.$