Question

Find the total number of nine digit numbers that can be formed using the digits $2,2,3,3,5,5,8,8,8$ so that the odd digit occupy the even places.

Answer 1

Places $\to \_\_\_\_\_\_\_\_\_$

So, No. of combinations $=4_{C_4}\times \frac{4!}{2!2!}\times 5_{C_5}\times \frac{5!}{2!3!}$

No. of ways $=\frac{4!\times 5!}{2!\times 2!\times 2!\times 3!}=\frac{24\times 120}{2\times 2\times 2\times 6}=60$ ways