Number of ways to choose $k$ balls such that there are at least $1$ of certain colors Suppose you have $n$ balls. Exactly $r$ of them are red, exactly $b$ of them are blue, and exactly $b'$ of the blues have a black dot on them. None of the red balls are blue, and none of the blue balls are red. The other balls are some other color.
You choose $k$ balls without replacement ($k$ is much less than $r$ or $b$, but it's greater than $b'$). How many ways are there to do each of the following?


*

*get $0$ red and $0$ blue

*get $0$ red, and at least $1$ blue without a black dot, and no blues with a black dot

*get $0$ red, and at least $1$ blue with a black dot

*get at least $1$ red, and $0$ blue

*get at least $1$ red, at least $1$ blue without a black dot, and no blues with a black dot

*get at least $1$ red, and at least $1$ blue with a black dot


Here is what I have so far on these numbers. I have numerically verified a few of them, and I'm fairly certain all of them are correct except for $(5)$ and $(6)$.


*

*$\displaystyle\binom{n-r-b}{k}$

*$\displaystyle\binom{n-r}{k} - \binom{n-r-(b-b')}{k}$

*$\displaystyle\binom{n-r}{k} - \binom{n-r-b'}{k}$

*$\displaystyle\binom{n-b}{k} - \binom{n-b-r}{k}$

*$\displaystyle\binom{n}{k} - \binom{n-r}{k} - \binom{n-(b-b')}{k} + \binom{n-r-(b-b')}{k}$

*$\displaystyle\binom{n}{k} - \binom{n-r}{k} - \binom{n-b'}{k} + \binom{n-r-b'}{k}$

 A: *

*Correct

*Wrong
Its ($0$ r and $0$ b') instead of ($0$ r)
therefore
$\displaystyle\binom{n-r-b'}{k} - \binom{n-r-b}{k}$

*Correct

*Correct

*Wrong
Sample space:$(n-b')$
To get atleast 1 red and 1 blue without a black dot
= Total - no red - no blue without a black dot
$\displaystyle\binom{n-b'}{k} - \binom{n-b'-r}{k} - \binom{n-b}{k} $

*Wrong
Sample space: $n$
To get atleast 1 red and 1 blue with a black dot
= Total - no red - no blue with a black dot
$\displaystyle\binom{n}{k} - \binom{n-r}{k} - \binom{n-b'}{k} $
Lets check the formula by following example
We have 4 red balls, 5 blue balls & 1 green ball 
total balls = 10
how many ways to choose 3 balls such that there is atleast 1 red ball and 1 blue ball
total no of ways to choose 3 balls out of 10 =$\binom{10}{3}$ =120
(RRR)=$\binom{4}{3}$=4
(RRB)=$\binom{4}{2}$$\binom{5}{1}$=30
(RBB)=$\binom{4}{1}$$\binom{5}{2}$=40
(BBB)=$\binom{5}{3}$=10
(RRG)=$\binom{4}{2}$$\binom{1}{1}$=6
(BBG)=$\binom{5}{2}$$\binom{1}{1}$=10
(RBG)=$\binom{4}{1}$$\binom{5}{1}$$\binom{1}{1}$=20
our desire cases=(RBG)(RRB)(RBB)
using above formula
Total - no red - no blue = Total- {(BBB)(BBG)(RRR)(RRG)} = (RBG)(RRB)(RBB)
