In how many ways can digits 0, 2, 4, 7 be arranged to make a 4 digit

Question : In how many ways can digits 0, 2, 4, 7 be arranged to make a 4 digit number without repetition

We have 4 distinct digits 0,2,4,7 . The different possible 4 digit numbers are

2047,2074,2470,2407,2704,2740
4027,4072,4207,4270,4702,4720
7024,7042,7204,7240,7402,7420 I.e 18 Numbers in all.

An easier way to solve this instead of writing down the numbers would be :

Lets say we have 2 digits say 1,2.
The number of 2 digits numbers possible is 2! ( 12,21)

Lets say we have 3 digits 0,1,2.
The possible number of 3 digits combinations possible is 3! ( 012,021,102,120,210,201 )
But 012 and 021 are not 3 digit numbers right ? They are 2 digit numbers ( 012 is same as 12, 021 is same as 21)
Hence the correct number of 3 digit numbers possible are 3! - 2 or (3! - 2!)

In the original question we have 4 distinct digits 0,2,4,7.
The possible no of 4 digit combination is 4! , but this includes combinations such as 0247,0274 etc,... which we need to subtract from 4!.
To find combinations such as 0274,0247.... we need to start the combination with 0 and the other 3 digits can be placed in any order.
0(2,4,7) - Number of ways to do this is 3!

Hence the num of 4 digits number possible is 4! - 3! = 18.

We don't have to subtract 2 digit or 1 digit number because we are not counting them.
Example : 2 digits number 24 or 47 are not part of our Initial possible set of numbers(4!) :

2047,2074,2470,2407,2704,2740
4027,4072,4207,4270,4702,4720
7024,7042,7204,7240,7402,7420
0247,0274,0427,0472,0724,0742

Hope this helps.