What is the probability of drawing a "conditionally" specific set of cards in a hand of 7? With the following specifics:


*

*Deck of 60 cards:
4 copies of $\mathcal A$,
3 copies of $\mathcal B$,
16 copies of $\mathcal C$ and
37 copies of $\mathcal Z$

*Hand of 7 cards drawn from the deck of 60, shuffled beforehand

*Order in the hand does not matter


1. Preliminary Problem
As a first step, I'm trying to express the probability of drawing a hand of $\mathcal A$, $\mathcal B$, $\mathcal C$, $\mathcal C$, $\mathcal Z$, $\mathcal Z$, $\mathcal Z$.
$$
P = \frac{\binom{4}{1}\binom{3}{1}\binom{16}{2}\binom{37}{3}}{\binom{60}{7}}
$$
If I was not to care about the 3 remaining cards (the 3 $\mathcal Z$) drawn in a way that they could be any combination of $\mathcal A$, $\mathcal B$, $\mathcal C$ and/or $\mathcal Z$, how should I fix my caclulation?
2. Follow-up Problem
Deck is now:
4 copies of $\mathcal A$,
3 copies of $\mathcal B$,
4 copies of $\mathcal C$,
12 copies of $\mathcal D$,
4 copies of $\mathcal E$,
15 copies of $\mathcal F$ and
18 copies of $\mathcal Z$
Now what if the valid cases I'm interested in are either:


*

*A hand of $\mathcal A$, $\mathcal B$, $\mathcal C$, $\mathcal C$, $\mathcal E$, $\mathcal ?$, $\mathcal ?$

*A hand of $\mathcal A$, $\mathcal B$, $\mathcal C$, $\mathcal C$, $\mathcal F
$, $\mathcal ?$, $\mathcal ?$

*A hand of $\mathcal A$, $\mathcal B$, $\mathcal C$, $\mathcal D$, $\mathcal E$, $\mathcal F$, $\mathcal ?$


Do I just sum these 3 probabilities together or is there a less verbose/slow known way of doing it? For context, I have small Python script that calculates probability of the hand seen in 1. and there are many more scenarii than just these 3.

Sketch:
Let the $4-$ tuple $(a,b,c,z)$ represent the number of cards of each type in your hand.  These are the patterns you wish to count.  Of course we must have $a+b+c+z=7$ but there are other restrictions as well.
The good hands satisfy:  $$4≥a≥1\quad 3≥b≥1\quad c≥2$$
We remark that we can drop the $4≥a$ restriction as it is implied by the others.
First step:  Just count the number of patterns.  Using the restrictions we see that we are trying to count the $4-$tuples $(a',b',c',z)$ of non-negative integers which add to $3$ and which satisfy $2≥b'$.  By Stars and Bars that count, without the $2≥b'$ constraint is given by $$\binom {7-1}3=20$$ substracting the single case with $b'=3$ we see that there are $19$ patterns to consider.  Not too bad.  we just have to enumerate these and count the ways to populate each. Defining $$a=a'+1\quad b=b'+1\quad c=c'+2$$ we remark that the number of ways to populate the pattern $(a,b,c,z)$ is $$\binom 4a\times \binom 3b\times \binom {16}c\times \binom {37}z$$
I don't see a quick way to simplify the sum.  Still, it's only $19$ terms.  It could be done by hand.
Added:  just did it by hand (mostly).  I got $$\boxed {19771136}$$  but I did it hastily and there could easily be blunders.
Remark:  might help an automated solution to note that defining $\binom nk=0$ when $k>n$ lets you ignore the cap on $b'$ (there would then be $0$ ways to populate that pattern).  For a more complex configuration this might simplify the coding.
