There are 15 balls in an urn. 5 red balls, 5 blue balls, and 5 white balls. Three balls are taken simultaneously.
The order of the balls does not matter so this is a combination problem. For each item, the sample space consists of
C315=15!3!(15−3)!=13×14×152×3=455C^{15}_3 = \frac{15!}{3!(15-3)!} = \frac{13 \times 14 \times 15}{2 \times 3} = 455C315=3!(15−3)!15!=2×313×14×15=455 items.
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