Consider a stack of $4n$ cards consisting of cards of all suits ($\color{red} \heartsuit$, $\color{red} \diamondsuit$, $\spadesuit$ and $\clubsuit$) of only numbered cards. Here are four separate problems: choose $5$ cards such that...
all are of the same suit
the five have consecutive numbers
exactly $3$ of which have the same number
not more than $2$ cards of each suit
Attempts: sub-part 1) $C(4n, 5)$ -> I am not able to evaluate it, although I do think this is the right way of going about it.
sub-part 2) Considering they are consecutive numbers, order does matter? I'm not sure how to proceed.
sub-part 3) Exactly $3$ cards have the same number, so $2$ cards are chosen randomly meaning $C(40,2)$ + number of ways $3$ cards of same number are chosen.
sub-part 4) $4 \cdot C(n-4, 2)$ ($4$ times you select $2$ cards randomly considering they are in the same suit)
Assuming there are exactly $n$ cards of each suit, here are some ideas.
Chose a suit ${4 \choose 1}$ them you have ${n \choose 5}$ possibilities. Multiply those numbers.
Sequences of five consecutive numbers must start with $2, 3, 4, 5$ or $6$ , so 5 possibilities.
Choose the repeating number ${9 \choose 1}$ (numbers form $2$ to $10$), then the remaining two, which cannot be equal to the number previously picked, but can also repeat.
There are two settings: all $4$ suits with one presenting $2$ cards, or $3$ suits total with two of them presenting $2$ cards. You will have to compute the possibilities for each setting individually.