Question

How many nine digit numbers can be formed using the digits $2$, $2$, $3$, $3$, $5$, $5$, $8$, $8$, $8$, so that the odd digits occupy even positions?

• A. $7560$
• B. $180$
• C. $16$
• D. $60$

Answer 1

The correct option is D $60$

Nine digit numbers are $2,2,3,3,5,5,8,8,8$

Here, the odd numbers as given are: $3355$

The even numbers are: $22888$

Hence, Number of Even places are $4$ and the number of Odd places are $5$

So we need to arrange $3355$ in the $4$ places by $\frac{4!}{(2!\times 2!)}$ ....(1)

and the even numbers in odd places by $\frac{5!}{(3!\times 2!)}$ .....(2)

From (1) and (2) $\frac{4!\times 5!}{(2!\times 2!\times 2!\times 3!)}=60$

Therefore there are $60$ possible ways.