Question

If a $6$ digits number's are formed using the digits $0,1,3,3,6,7$ and arranged in ascending order, then the position of the number ${}^{\prime \prime }{631307}^{\prime \prime }$ is

Answer 1

Numbers cannot start from $0$
Total numbers starting from $1$ are $\frac{5!}{2!}=60$
Total numbers starting from $3$ are $5!=120$
Total numbers starting from $60$ are $\frac{4!}{2!}=12$
Total numbers starting from $61$ are $\frac{4!}{2!}=12$
Total numbers starting from $630$ are $3!=6$
Total numbers starting from $6310$ are $2!=2$
Next number will be ${}^{\prime \prime }{631307}^{\prime \prime }$
position of the number ${}^{\prime \prime }{631307}^{\prime \prime }=60+120+12+12+6+2+1=213$