Answer
$10$
Work Step by Step
The number of three-letter (unordered) sets that use letters b,o,g,e,y at most once each equals to the number of combinations of 5 items taken 3 at a time : $C(5,3)=\frac{5!}{3!(5-3)!}=\frac{5!}{3!*2!}=\frac{5*4*3*2*1}{3*2*1*2*1}=5*2=10$