(c): Four digits $3,3,5,5$ can be arranged at four even places in $\frac{4!}{2!2!}=6$ ways. $\\$ Remaining digits $2,2,8,8,8$ can be arranged at the five odd places in $\frac{5!}{2!3!}=10$ ways. $\\$Hence number of possible arrangements $=6 \times 10=60$.