Question

There are 15 balls in an urn. 5 red balls, 5 blue balls, and 5 white balls. Three balls are taken simultaneously.

Answer 1

The order of the balls does not matter so this is a combination problem. For each item, the sample space consists of

$C^{15}_3 = \frac{15!}{3!(15-3)!} = \frac{13 \times 14 \times 15}{2 \times 3} = 455$ items.