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.
Corrected: Thanks to the OP for querying my previous answer, and to joriki for pointing out that I was counting the wrong thing.
There are $\binom{52}{13}$ ways of choosing $13$ cards, all equally likely.
We will be finished if we count the number of hands that have exactly one void.
This number is $4$ times the number of hands void in $\spadesuit$ only.We now proceed to count these.
The number of hands void in $\spadesuit$ is $\binom{39}{13}$. This overcounts the hands void in $\spadesuit$ alone. To adjust, we use the
Inclusion-Exclusion Principle.
How many hands are void in both $\spadesuit$ and $\heartsuit$? Clearly $\binom{26}{13}$. The same is true for $\spadesuit$ and $\diamondsuit$, and for $\spadesuit$ and $\clubsuit$. So from our first estimate of $\binom{39}{13}$ we subtract $3\binom{26}{13}$.
But we have subtracted too much. We need to add back the number of hands that are void in all but one of $\heartsuit$, $\diamondsuit$, or $\clubsuit$. There are $3$ of these. Thus the number of hands with exactly one void is
$$4\left(\binom{39}{13}-3\binom{26}{13}+3\right).$$
Comment: From the "practical" point of view, we could have stopped with the first term, since in a well-shuffled deck multiple voids have negligibly small probability compared to single voids.
Best Answer
Include the number of ways to choose from all the cards except for $1$ out of $4$ suits:
$$\binom{4}{1}\cdot\binom{64}{18}$$
Exclude the number of ways to choose from all the cards except for $2$ out of $4$ suits:
$$\binom{4}{2}\cdot\binom{50}{18}$$
Include the number of ways to choose from all the cards except for $3$ out of $4$ suits:
$$\binom{4}{3}\cdot\binom{36}{18}$$
Exclude the number of ways to choose from all the cards except for $4$ out of $4$ suits:
$$\binom{4}{4}\cdot\binom{22}{18}$$
Divide the result by the number of ways to choose from all the cards:
$$\frac{\binom{4}{1}\cdot\binom{64}{18}-\binom{4}{2}\cdot\binom{50}{18}+\binom{4}{3}\cdot\binom{36}{18}-\binom{4}{4}\cdot\binom{22}{18}}{\binom{78}{18}}\approx6.72\%$$