Answer
$20$
Work Step by Step
The number of three-letter (unordered) sets that use letters q,u,a,k,e,s at most once each equals to the number of combinations of 6 items taken 3 at a time : $C(6,3)=\frac{6!}{3!(6-3)!}=\frac{6!}{3!*3!}=\frac{6*5*4*3*2*1}{3*2*1*3*2*1}=5*4=20$