Q:

Selecting two types of objects from a group with two types of objects

I have two types of objects: RED objects (R) and BLUE objects (B). I have $X$ total red objects and $Y$ total blue objects. What's the probability of drawing $a$ red objects and $b$ blue objects, for $0 <= a <= X$ and $0 <= b <= Y$?
I'm struggling with this question, which is related to deck draws in a certain social deduction game, mainly because I'm uncertain of what the combinatorials should be.
I've tried $$\frac{\binom{X}{a}}{\binom{X+Y}{a}} \cdot \frac{\binom{Y}{b}}{\binom{X+Y}{b}}$$
But this produces answers that are incorrect. My verifier of correctness is a spreadsheet with all the values that may be incorrect itself. Should this be giving me the right answer? If not, what would?
NOTE: Order of drawing the RED and BLUE objects doesn't matter: so 2R 1B is the same as 1R 1B 1R, for example.

A:

Total number of outcomes when drawing $a+b$ objects from $X+Y$ total objects is $N=\binom{X+Y}{a+b}$. These outcomes are equally likely, so probability of any event  $A$ equals to 
$$
Prob(A) = \dfrac{\text{number of outcomes in event } A}{\text{total number of outcomes}}
$$
Number of drawings when $a$ red objects and $b$ blue objects are obtained is $n=\binom{X}{a}\binom{Y}{b}$. 
Probability of this event is
$$
Prob(a \text{ red}, b \text{ blue}) = \frac{n}{N} = \frac{\binom{X}{a}\binom{Y}{b}}{\binom{X+Y}{a+b}}.
$$
Look also at https://en.wikipedia.org/wiki/Hypergeometric_distribution 

