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.
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.
Best Answer
Let $C(n,k)$ be the binomial coefficient, then the probability of a particular suit NOT appearing is $$\frac{{C(16 - 4,8)}}{{C(16,8)}}.$$ Similarly, the probability of two suits NOT appearing is $$\frac{{C(16 - 8,8)}}{{C(16,8)}}$$ and the probability of three suits NOT appearing is $$\frac{{C(16 - 12,8)}}{{C(16,8)}}$$ and the probability of four suits not appearing is $$\frac{{C(16 - 16,8)}}{{C(16,8)}} = 0.$$ Therefore, the probability that at least one of the suits does NOT appear is $$\frac{{4C(16 - 4,8) - 6C(16 - 8,8) + 4C(16 - 12,8)}}{{C(16,8)}}$$ so $$1 - \frac{{4C(16 - 4,8) - 6C(16 - 8,8) + 4C(16 - 12,8)}}{{C(16,8)}}$$ is the probability we seek (4,-6, and 4 come from the inclusion-exclusion principle). Watch this YouTube video if you do not understand the inclusion-exclusion principle.