# Inclusion-Exclusion Question with “at least”

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.

Thanks!

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