Chapter 6 - Section 6.4 - Permutations and Combinations - Exercises - Page 437: 27

$60$

First we choose where to put the k in the six-letter sequence. That is 6 choices. Then we choose where to put the two u-s ( 5 places left). That is $C(5,2)=\frac{5!}{2!(5-2)!}=\frac{5!}{2!*3!}=\frac{5*4*3*2*1}{2*1*3*2*1}=\frac{5*4}{2}=10$ The other letters are a's. This gives a total of $6*10=60$ different six-letter sequences that use the letters a,u,a,a,u,k