Question

How many numbers between 2000 and 3000 can be formed from the digits 2, 3, 4, 5, 6, 7 when repetition of digits is not allowed ?

Answer 1

Clearly, a number between 2000 and 3000 will have 2 at the thousand's place. So, this place can be filled in 1 way only.

Now, the hundred's place can be filled by any of the remaining five digits. So, there are 5 ways of filling the hundred's place.

Similarly, the ten's place can be filled in 4 ways and the unit's place can be filled in 3 ways.

Hence, by the fundamental principle of multiplication, the total number of required numbers = $(1\times 5\times 4\times 3)=60.$