Making sense of the binomial factor. 
Five cards are drawn at random from a deck of cards. Let $X$ be the
  number of aces. Find the pmf of $X$ if the cards are drawn with replacement.

The answer is $$p(k)={5\choose k}\left(\frac{1}{13}\right)^k\left(\frac{12}{13}\right)^{5-k}.$$ 
I understand the last two factors, but I don't have any intuitive sense as to why we need the binomal there. I understand that this follows directly from the binomal distribution formula and so on but I want to know how to intuitively deduct the answer.
 A: Out of $5$ cards there must be $k$ of them aces, so ${5\choose k}$.

If it would be without replacement then it would be $p(k)=0$ for $k\geq 2$, $p(1) = {1\over 13}$ and $p(0) = {12\over 13}$. 
A: We can view the experiment of drawing $5$ aces as $5$ Bernoulli trials, and count it as a success if the drawn card is an ace. The number of successes (because we are allowing repetition) can be $0,1,2,3,4$ or $5$. 
Now, $n$ trials can result in $k$ successes (and hence, $n-k$ failures) in the same number of ways that $k$ letters can be distributed among $n$ places. This is, in $\binom{n}{k}$ ways.
For example: suppose you want to calculate $P(X=4)$. Then you have that
$$P(X=4)= \binom{5}{4}\left(\frac{1}{13}\right)^4\left(\frac{12}{13}\right)^{1}$$
Why? because you need $4$ successes, each with probability $1/13$, hence the middle term. You also need a failure, with probability$12/13,$ hence the last term.  But notice that all of the following  are favourable outcomes of the experiment:
$$AAAAx,\ AAAxA,\ AAxAA, \ AxAAA, \ xAAAA$$
where $A $ is an ace and $x$ any other card. Therefore we have to consider that our $k$ successes can occur in any order in the $n$ trials
