Question

How many numbers can be formed by using all the digits $1,2,3,4,3,2,1$ so that the odd digits always occupy the odd places?

• A. $16$
• B. $17$
• C. $18$
• D. $20$

Answer 1

Answer: (C)

$18$

Out of these $1,3,1$ and$3$ are odd digits for them to occupy the odd, places there are $\frac{4!}{2!.2!}$ ways = $6$ ways
and
$2,4$ and$2$ are even digits for them to occupy the even places there are
$\frac{3!}{2!.1!}$ ways = $3$ ways
So, there are $6\times 3$ ways to form a number= $18$ ways
Hence, option 'C' is correct.