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

Answer: (D)

$60$

We have $4$ even placed and $5$ odd place.

Odd digits are $3355$

They have to be placed in $4$ even places so no. of ways $\frac{4!}{2!2!}$

and even digits are $22$ no. of ways $\frac{5!}{3!2!}$

Total $\frac{4!}{2!2!}\times \frac{5!}{3!2!}$

$=60$ ways.

Hence, the answer is $60.$