Answer
$60$
Work Step by Step
The number of three-letter sequences that use the letters b,o,g,e,y at most once each equals to the number of permutations of 5 items taken 3 at a time : $P(5,3)=\frac{5!}{(5-3)!}=\frac{5!}{2!}=\frac{5*4*3*2*1}{2*1}=5*4*3=60$
