A meal comes with two side dishes chosen from five options, and the order of the sides does not matter. How many selections are possible?

1. Order does not matter, so this is a combination of 5 objects taken 2 at a time. 2. Apply C(n, r) = n!/(r!(n - r)!) with n = 5 and r = 2. 3. That gives 5!/(2! times 3!) = 120/(2 times 6) = 120/12. 4. So there are 10 selections. 5. Counting ordered pairs would give 5 times 4 = 20, which lists each pair twice. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 13 "Sequences, Probability and Counting Theory", section 13.5 Counting Principles_