For the following question:

Choose 5 balls from 6 blue balls and 10 red balls. How many ways to choose at least 3 blue balls?

I can solve this by counting when having exactly 3, 4, or 5 blues balls to get the answer. But I'm having a hard time to use inclusion-exclusion to solve this. When I have

$\binom{6}{3}\cdot\binom{16 - 3}{2}$ I overcounted. But I'm not sure how to proceed from here.


  • $\begingroup$ P(1)-P(2)+P(3) = at least 3 $\endgroup$ – Saketh Malyala Apr 23 '17 at 6:32
  • $\begingroup$ What is P(1)? at least 1 blue? $\endgroup$ – jiangp97 Apr 23 '17 at 6:40
  • $\begingroup$ Are the blue balls distinguishable, or identical? the red balls? $\endgroup$ – Gerry Myerson Apr 23 '17 at 9:22

You are insterested in $3$ events. $3$ or $4$ or $5$ blue are chosen. $\cup$ of events ORs them, and the principle can be used.

  • $\begingroup$ For event "3", you mean exactly 3 are chosen? $\endgroup$ – jiangp97 Apr 23 '17 at 6:41
  • $\begingroup$ Yes exactly 3 are chosen $\endgroup$ – user76568 Apr 23 '17 at 6:45
  • $\begingroup$ Okay cool. I get this way. I guess I'm more curious about how to subtract from the equation I started using inclusion-exclusion $\endgroup$ – jiangp97 Apr 23 '17 at 6:46

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