Answer
$120$
Work Step by Step
The number of three-letter sequences that use the letters q,u,a,k,e,s at most once each equals to the number of permutations of 6 items taken 3 at a time : $P(6,3)=\frac{6!}{(6-3)!}=\frac{6!}{3!}=\frac{6*5*4*3*2*1}{3*2*1}=6*5*4=120$
