first draw: Pick any card, probabilty 1 you are still OK
second draw: you must pick from 39 cards that won't wreck your hand out of 51 cards
third draw: you must pick from 26 of the remaining 50
fourth draw: you mustpick from 13 of the remaining 49.
Altogether, you get a probability of
$$1\cdot {39\over 51}\cdot{26\over 50}\cdot{13\over 49}. $$
You have it.
Here is a second solution. There are ${52\choose 4}$ hands of size 4.
Now pick the four cards of different suits; there are $13^4$ ways to do
this.
When $5$ cards are dealt, the reasonable assumption is that they are dealt from the same deck, so the number of hands is $\dbinom{52}{5}$. All these hands are equally likely.
Now let us count how many hands have at least one card from each suit.
The suit that we have $2$ of can be chosen in $\dbinom{4}{1}$ ways. For each of these ways, the actual $2$ cards can be chosen in $\dbinom{13}{2}$ ways. For each of these choices, there are $\dbinom{13}{1}$ ways to choose $1$ card from the highest ranking remaining suit, and so on, for a total of
$$\binom{4}{1}\binom{13}{2}\binom{13}{1}\binom{13}{1}\binom{13}{1}.$$
To find the probability, divide.
Now we can look at an entirely different problem, in which after drawing a card and writing down what it is, we replace the card in the deck and shuffle before drawing again. That is a somewhat unreasonable interpretation of the problem.
The natural denominator is then $52^5$, since there are $52^5$ sequences of cards, all equally likely. Now we need to count the number of "favourables."
The suit we will have $2$ of can as before be chosen in $\dbinom{4}{1}$ ways. For each of these ways, the location of the draws at which we got this suit can be chosen in $\dbinom{5}{2}$ ways. For each way of specifying the location, the two slots can be filled in $13^2$ ways. Once this is done, there are $3$ empty slots. The first one can be filled in $39$ ways. For each such way, the next empty slot can be filled in $26$ ways. And for each of these, the remaining empty slot can be filled in $13$ ways.
There are other ways of organizing the count. For example, onece we are left with $3$ empty slots, there are $3!$ ways of deciding the order of the suits in thse empty slots. And once this is done, the slots can be filled in $(13)(13)(13)$ ways.
Best Answer
You are double counting.
How many ways are there to get at least six hearts? That's $6$ up to $13$ hearst and then the rest are non-hearts, i.e., $${13\choose 6}{39\choose 7}+{13\choose 7}{39\choose 6}+{13\choose 8}{39\choose 5}+{13\choose 9}{39\choose 4}+{13\choose 10}{39\choose 3}+{13\choose 11}{39\choose 2}+{13\choose 12}{39\choose 1}+{13\choose 13}{39\choose 0}$$ The same goes for the other three suits. However, this double-counts all possibilities where more than one suit has at least six cards! Fortunately, it is not possible that three suits have six or more carsd among 13, so we only count the ways to have at least six hearts and at least six spades, say: $${13\choose 6}{13\choose 6}{26\choose 1}+{13\choose 6}{13\choose 7}{26\choose 0}+{13\choose 6}{13\choose 7}{26\choose 0} $$ Now to arrive at the correct count for the problem, multiply the result from the first formula by four (for the four suits), multiply the result from the second formula by six (for the six possible pairs of distinct suits) and subtract the latter from the former. Of course, ultimately divide by $52\choose 13$ to arrive at a probability.