If the letter z's are distinctly different, there are P(5,5) arrangements. That is
[tex]P\left(5,5\right)=5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120[/tex]
If the z's are interchangeable, the number of different arrangements is given by
[tex]\frac{5!}{2!}=5\cdot 4\cdot 3=60[/tex]
