Question

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

• A. $18$
• B. $26$
• C. $6$
• D. $3$

Answer 1

Answer: (A)

Given digits $1,2,3,4,3,2,1$
We need, $7$ 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 $1’s$ and two $3’s$.
$\therefore 4$ odd digits can be arranged at $4$ odd places in $\frac{{}^4{P}_4}{2!2!}$
$=6$ ways.
Now, there are $3$ even places.
$\therefore 3$ even digits can be arranged in $\frac{3!}{2!}$ ways. (As there is two $2^{\prime }s$)
$\therefore$ Total number of ways are $6\times 3=18$