Question

How many valid 3 digit numbers can you make using the digits 0, 1, 2 and 3 without
repeating the digits? How about with repeating?

Answer 1

I)Digits cannot be repeated

Solution: There are 4-digits :0,1,2,3,

The digits to be formed =No.of places=3

(I)Case I: Digits cannot be repeated:If 0 is placed in first place then it becomes a 2-digit number out of 3-places.Thus ,we can fill 1 or 2 or 3 in the first place.

Therefore,No.of possibilities in the first place =3

Again,consider the second place.Here we can fill 0 and any of the 2 digits

Thus, No.of possibilities=3 (the digit 0 and 2 digits)

Consider the third place.We can fill any of the 2 digits.

Thus, No.of possibilities=2

We arrange the numbers as :- 3 x 3 x 2 = 18

Case 2) Repetition is allowed

For the first digit, we can choose only from 3 out of 4 digits , as one of the digits is zero(0).

For the second digit, we can choose any one from all 4 digits.

Similarly, for the third digit, we can again choose any one out of given 4 digits.

Therefore , total number of 3 digits numbers = 3*4*4 = 48