combinatorics: deck of cards and suits I feel confused about selecting suits and cards that each suit corresponds to.
Suppose we have a standard deck of cards, and we want to form a 6 hands with at most 3 suits.
I understand this needs to be spliced into 3 cases to discuss, which are 1 suit case, 2 suits case, and 3 suits case.
But I feel confused when it comes to select cards.
For example, I am not sure if I understand this correctly, but I feel like if we only need to form cards with a single suit, then this can be done by $4(13 * 12 * 11 * 10 * 9 * 8)$ ways, since each suit has 13 cards, and we have 4 suits here.
And when it comes to the 2 suits case, I feel like we only need to draw from the total of 26 cards to form such hands, and there are 6 ways to do this, but I am not sure if my understanding about the 26 cards is right or what.
And for the 3 suits case, I feel like it's just choose 3 suits out of 4, and draw cards from total of 39 cards.
Is my understanding towards the total number of cards in each case correct? If not, why so?
 A: There is no need to approach by breaking into cases.  Case work will just become too messy and does not generalize nicely to other sized hands.
Consider answering the question instead as: "How many ways are there to form a six-card hand such that at least one of the suits is missing?"
Let $A$ be the event that clubs are missing, $B$ that diamonds are missing, $C$ that hearts are missing, and $D$ that spades are missing respectively.  You are tasked with finding $\Pr(A\cup B\cup C\cup D)$
This expands by inclusion exclusion to be:
$$\Pr(A\cup B\cup C\cup D) = \Pr(A)+\Pr(B)+\Pr(C)+\Pr(D)-\Pr(A\cap B)-\Pr(A\cap C)-\dots - \Pr(C\cap D) + \Pr(A\cap B\cap C)+\dots + \Pr(B\cap C\cap D) - \Pr(A\cap B\cap C\cap D)$$
By symmetry, this simplifies to be:
$$ = 4\cdot \Pr(A) - 6\cdot \Pr(A\cap B) + 4\cdot \Pr(A\cap B\cap C)$$
noting that we could leave off the intersection of all four since it is impossible for all suits to be missing simultaneously.
The probability that clubs are missing?  That is answered simply by a hypergeometric distribution argument and is $\Pr(A) = \dfrac{\binom{39}{6}}{\binom{52}{6}}$.  Similarly for the probability that both clubs and diamonds are missing as $\dfrac{\binom{26}{6}}{\binom{52}{6}}$ and so on.
Your probability is then:
$$\dfrac{4\cdot\binom{39}{6} - 6\cdot\binom{26}{6}+4\cdot\binom{13}{6}}{\binom{52}{6}}$$
