Question

Different $7$ digit numbers that can be formed using digits $1,2,3,3,4,4,6$, such that the number formed is divisible by $2$ and all the prime numbers always occupies the even places only is

Answer 1

Given digits are 1,2,3,3,4,4,6 so starting from right
Prime number $i.e.2,3,3$ will occupy only $2^{rd},4^{th},\&6^{th}$ places
$\because$ number is divisible by $2$
$\therefore$ last digit can be $4\text{or}6$
Case-I: If last digit is $4$ then rest of the $3$ places can be filled by $3!$ ways.
Case-II: If last digit is $6$ then rest of the $3$ places can be filled by $\frac{3!}{2!}$ ways.
So total different $7$ digit number is =$\frac{3!}{2!}(\frac{3!}{2!}+3!)=27$