I have a feeling your teacher/professor intended for you to learn the following (we first introduce some notation):
For two sets $A, B$ in a common universe $U$, define their union as
$$
A \cup B = \{x \in U : x \in A \text { or } x \in B\}.
$$
Define their intersection as:
$$
A \cap B = \{x \in U : x \in A \text{ and } x \in B\}.
$$
It is important not to get too wrapped in the English here. Being in $A \cup B$ simply means being in one of the two sets (or possibly both). Being in $A \cap B$ simply means being in set $A$ and being in set $B$ at the same time.
Finally, a finite set $A$ with $k$ elements ($k$ things in the set) has cardinality $k$, and this is written as $|A| = k$, or $\#A = k$ or even sometimes, $n(A) = k$.
Therefore:
$$
\{1,3,5\} \cup \{1,2,3\} = \{1,2,3,5\},
$$
while
$$
\{1 ,3 , 5 \} \cap \{ 1 , 2, 3 \} = \{ 1, 3\}.
$$
Also,
$$
| \{ 1,3,5\}| = 3,
$$
while
$$
| \{ 1, 3\} | = 2.
$$
Now, what your teacher probably wanted you to learn was the following "rule":
$$
|A \cup B| = |A| + |B| - |A \cap B|.
$$
This is easy to see it is true, since to count the number of elements that are in either $A$ or $B$ (or possibly both $A$ and $B$), you count the number of elements in $A$, add to it the number of elements in $B$, and then subtract the stuff you double counted, which is precisely the elements in $A \cap B$.
Therefore, if you want to find the number of sides of a die that are (say) even or prime, you count the number of sides which are even (there are $3$ such sides - namely $2$ , $4$, $6$) add to it the number of sides which are prime (again, there are $3$ such sides - namely $2$, $3$, $5$), and then subtract the sides which we double counted (we counted the side with the number $2$ twice).
Therefore, there are $3 + 3 - 1 = 5$ sides of a die which are even or prime.
Now, you can take this strategy and count the number of cards in a deck which are either non face cards or clubs.
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
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.