Question

The number of different nine digit numbers that can be formed from $22,33,55,888$ by rearranging the digits, so that odd digits occupy even places and even digits occupy odd position, is

• A. $16$
• B. $60$
• C. $36$
• D. $180$

Answer 1

Answer: (B)

$60$

There are $4$ even places and $5$ odd places.

Odd digits are $3355$.

Odd numbers are placed in these $4$ places in $\frac{4!2!}{2!}$ ways.

Even digits are $22888$.

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

$\therefore$ No. of required $9$ digit numbers$=\frac{4!}{2!.2!}.\frac{5!}{3!.2!}$

$=\frac{4\times 3\times 2\times 1}{2\times 2}\times \frac{5\times 4\times 3\times 2\times 1}{3\times 2\times 2}=60$