Question

Number which can be formed with digits 1, 2, 3, 4, 3, 2, 1 using all digits so that odd digits occupy always the odd place are

• A. $6$
• B. $12$
• C. $15$
• D. $18$

Answer 1

The correct option is D $18$

$1,2,3,4,3,2,1$

number of odd digits is $4$

no of odd places - $4$

number of ways of arranging = $\frac{4!}{2!\times 2!}=6$

number of places left = $3$

possible arrangements = $\frac{3!}{2!}=3$

total number of ways = $6\times 3=18$ ways