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

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  • $\begingroup$ I think 5 and 6 are correct. OP applied inclusion exclusion principle correctly I guess $\endgroup$ – Rohan Shinde Mar 4 '18 at 6:12
  • $\begingroup$ Earlier i also thought that the OP applied correctly but when i used the test case it did not come out true.Possibly because there is no common cases that is double counted. I have wrote the test case, Hopefully it helps $\endgroup$ – NewGuy Mar 4 '18 at 7:37
  • $\begingroup$ I don't think your #5 is correct. I tried it with n=10, r=2, b=3, b'=1, and k=4, and I got 61. RRBB = 1 RRBG = (2 choose 1)*(5 choose 1) = 10 RBBG = 10 RBGG = 40 Total: 61 $\endgroup$ – Nathan H. Mar 11 '18 at 17:35
  • $\begingroup$ @Nathan H its given in the question that k is less than r or b $\endgroup$ – NewGuy Mar 11 '18 at 18:22
  • $\begingroup$ Yes, you are right. Forgot about that. I'll try it with a smaller k and see if it works. Also, could you explain why my #2 is wrong? I drew a Venn diagram to see how to write the formula, and I don't see how my formula doesn't give the right answer. What is my formula counting? $\endgroup$ – Nathan H. Mar 11 '18 at 21:59

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