8 cards are drawn randomly from a pack of cards (52) with no returns.
We define these events:
S - not getting spade in those 8 cards
H - not getting heart in those 8 cards
D - '' '' diamond '' '' ''
C - '' '' Clubs '' '' ''
What is the probability at least one card from each series is in the cards chosen (e.g. there is at least one heart, at least one club at least one spade and at least one diamond)?
The probability we are interested in is the probability that in our hand we do have at least one spade and we do have at least one heart as well as at least one club and at least one diamond.
If we were to let $S,H,C,D$ be as defined above as the events that we do not have a card of the corresponding type in our hand, we are wanting to calculate $Pr(S^c\cap H^c\cap C^c\cap D^c)$, i.e. the probability that it is not the case that we have no spades as well as it is not the case that we have no hearts, etc...
So... to approach, we do this by looking at the opposite event. $(S^c\cap H^c\cap C^c\cap D^c)^c$ which simplifies as $S\cup H\cup C\cup D$, seen by De Morgan's Laws.
Now, the probability $Pr(S\cup H\cup C\cup D)$ can be calculated via inclusion-exclusion as $$Pr(S)+Pr(H)+Pr(C)+Pr(D)-Pr(S\cap H)-Pr(S\cap C)\dots - Pr(C\cap D) + Pr(S\cap H\cap C)+\dots Pr(H\cap C\cap D) - Pr(S\cap H\cap C\cap D)$$
Recognizing the symmetry of the problem, that $Pr(S)=Pr(H)=Pr(C)=Pr(D)$ and $Pr(S\cap H)=Pr(S\cap C)=\dots = Pr(C\cap D)$ and so on... and that there are $4$ ways to write a single set by itself, $\binom{4}{2}=6$ ways to write two sets intersected, $\binom{4}{3}=4$ ways to write the intersection of three sets, etc... this simplifies further as:
$$4\cdot Pr(S)-6\cdot Pr(S\cap H)+4\cdot Pr(S\cap H\cap C) - Pr(S\cap H\cap C\cap D)$$
Now... to calculate the probability $Pr(S)$, this is the probability that among our eight cards, none of them were spades. There are $39$ non-spade cards of which we want to pick eight of them to be used in our hand. Dividing by $\binom{52}{8}$, the number of eight-card hands possible, and noting that each of those eight-card hands are equally likely to occur, gives us the probability $$Pr(S)=\dfrac{\binom{39}{8}}{\binom{52}{8}}$$
Similarly, $Pr(S\cap H)$ is the probability that among our eight cards none of them were hearts as well as none of them were spades. There are $26$ non-heart non-spade cards of which we choose eight of them... giving $$Pr(S\cap H) = \dfrac{\binom{26}{8}}{\binom{52}{8}}$$
Continue in this fashion and recognize that it is impossible to have none of any of hearts spades clubs diamonds since having at least one card in hand automatically implies that we have at least one of those suits present, so the final term was zero and can be left out.
This all together gives us:
$$Pr(S\cup H\cup C\cup D) = \dfrac{4\binom{39}{8}-6\binom{26}{8}+4\binom{13}{8}}{\binom{52}{8}}$$
Remember though that $Pr(S\cup H\cup C\cup D)$ is not the actual probability we are interested in, but rather is the opposite probability of what we are interested in. The probability we were actually interested in is:
$$Pr(S^c\cap H^c\cap C^c\cap D^c) = 1 - Pr(S\cup H\cup C\cup D) = 1 - \dfrac{4\binom{39}{8}-6\binom{26}{8}+4\binom{13}{8}}{\binom{52}{8}}$$