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

Given digit are $2,3,3,5,5,8,8,8$

There are 4 even places and 5 odd places odd digits are $3,3,5,5$

odd number are placed in these 4 places in $\frac{4!}{2!2!}$ ways

even digits are $2,2,8,8,8$

These even digits are placed in 5 odd places in $\frac{5!}{3!2!}$ ways

$\therefore$ No. of required a digit numbers are $\frac{4!}{2!2!}\cdot \frac{5!}{3!2!}$

$=\frac{4\times 3!\times 5\times 4\times 3\times 2\times 1}{2\times 2\times 3!\times 2}$

$=60$