Question

Given the digits $2,4,6$ and $9$, how many $4$ digit numbers can be formed if the numbers have different digits and are greater than $4000$?

Answer 1

A $4$ dit numbers has $4$ places for digits to be filled.

At thousand's place we can insert $4,6,9$. So, we have $3$ choices. Now, for hundred's place, we have $3$ choices only.

Similarly, for tenth and unit place, we have $2$ and $1$ choices respectively.

The choices are getting reduced because we cannot repeat digits after inserting them.

Thus, the number of ways are $=3\times 3\times 2\times 1=18$.

Hence, total number of $4$ digits numbers are $18$.