Question

The number of different numbers that can be formed by using all the digits $1,2,3,4,3,2,1$, so that odd digits always occupy the odd places is

• A. 6
• B. 72
• C. 60
• D. 18

Answer 1

The correct option is D 18

One of the possible arrangements is $1$ $2$ $3$ $4$ $3$ $2$ $1$

Here, the odd places are filled as $1$ $3$ $3$ $1$

These numbers can be arranged among themselves in $\frac{4!}{2!\times 2!}$ ways

The even places are filled as $2$ $4$ $2$

These numbers can be arranged among themselves in $\frac{3!}{2!}$ ways

Hence, the total number of ways is

$\frac{4!}{2!\times 2!}\times \frac{3!}{2!}=18$

Answer is Option D