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.
Hints:
Combinatorial way 1: There are $\binom{52}{4}$ equally likely hands of $4$. There are $\binom{39}{4}$ no club hands.
Combinatorial way 2: Imagine dealing the cards one at a time. There are $(52)(51)(50)(49)$ equally likely strings of four distinct cards. How many strings of $4$ cards have no club?
Conditional probability way: Imagine that the cards are dealt one at a time. The probability the first card is not a club is $\frac{39}{52}$.
Given that the first card is not a club, the probability the second card is not a club is $\frac{38}{51}$. So the probability that neither of the first two cards is a club is $\frac{39}{52}\cdot \frac{38}{51}$.
Given that neither of the first two card is a club, the probability the third card is not a club is $\frac{37}{50}$. Continue.
Best Answer
Let $s=13$ denote the number of cards in each suit and $N$ denote the (random) number of cards drawn when all $k=4$ suits are represented for the first time. Hence, $k\leqslant N\leqslant(k-1)s+1$ with full probability.
For each $n\geqslant1$, the event $[N\gt n]$ depends on the $n$ first cards drawn only. There are ${ks\choose n}$ collections of $n$ cards from a full deck of $ks$ cards and each such collection has the same probability ${ks\choose n}^{-1}$ to be drawn. The event $[N\gt n]$ means that one avoids at least one suit. Using inclusion-exclusion principle, there are $A_n$ ways to do so, where $$ A_n={k\choose1}{(k-1)s\choose n}-{k\choose2}{(k-2)s\choose n}+\cdots\pm{k\choose k-1}{s\choose n}\mp{k\choose 0}{0\choose n}. $$ This yields the expectation $$ \mathbb E(N)=\sum_{n\geqslant0}\mathbb P(N\gt n)=\sum_{n\geqslant0}{ks\choose n}^{-1}A_n. $$ In the case at hand, $$ \mathbb E(N)=\sum_{n\geqslant0}{52\choose n}^{-1}\left(4{39\choose n}-6{26\choose n}+4{13\choose n}-{0\choose n}\right), $$ that is, $$ \mathbb E(N)=4+\sum_{n=4}^{39}{52\choose n}^{-1}\left(4{39\choose n}-6{26\choose n}+4{13\choose n}\right), $$ or, $$ \mathbb E(N)=4+4B_{39}-6B_{26}+4B_{13},\qquad B_i=\sum_{n=4}^{i}{52\choose n}^{-1}{i\choose n}. $$ Numerically, $$ \mathbb E(N)=\frac{4829}{630}=7+\frac23-\frac1{630}\approx7.66508. $$