Question

Number of numbers that can be formed by using all the digits 1, 2, 3, 4, 3, 2, 1 so that odd digit always occupy odd places

• A. $18$
• B. $26$
• C. $6$
• D. $3$

Answer 1

The correct option is A $18$

Given digits $1,2,3,4,3,2,1$

We needed $7$ digit number, which is same as filling $7$ places in a row with the $7$ digits $1,2,3,4,3,2,1$

Odd places are to be filled by $1,3,5,7$ digits.

Odd digits are available $1,3,3,1$

i.e. two is $1^{\prime }s$ and two. $3^{\prime }s$.

$4$ odd digit can be arranged at $4$ odd places in $\frac{{}^4{P}_4}{2!2!}=6$ ways.

Now there are $3$ even places.

Required such number $=\frac{3!}{2!}$

Now using fundamental Law of multiplication.

Total number of ways are $6\times 3=18$