Inclusion-Exclusion Question with "at least"

Question

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!

Answer 1

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.