Number of ways to choose $k$ balls such that there are at least $1$ of certain colors

Question

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?

1. get $0$ red and $0$ blue
2. get $0$ red, and at least $1$ blue without a black dot, and no blues with a black dot
3. get $0$ red, and at least $1$ blue with a black dot
4. get at least $1$ red, and $0$ blue
5. get at least $1$ red, at least $1$ blue without a black dot, and no blues with a black dot
6. 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 $(6)$.

1. $\displaystyle\binom{n-r-b}{k}$
2. $\displaystyle\binom{n-r}{k} - \binom{n-r-(b-b')}{k}$
3. $\displaystyle\binom{n-r}{k} - \binom{n-r-b'}{k}$
4. $\displaystyle\binom{n-b}{k} - \binom{n-b-r}{k}$
5. $\displaystyle\binom{n}{k} - \binom{n-r}{k} - \binom{n-(b-b')}{k} + \binom{n-r-(b-b')}{k}$
6. $\displaystyle\binom{n}{k} - \binom{n-r}{k} - \binom{n-b'}{k} + \binom{n-r-b'}{k}$

Answer 1

1. Correct

2. Wrong

   Its ($0$ r and $0$ b') instead of ($0$ r)

   therefore

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

3. Correct

4. Correct

5. 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} $

6. 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)