Question

How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3 and 4, if the digits can repeat?

Answer 1

Case I: Four-digit number
Total number of ways in which the 4 digit number can be formed = 4$\times$4$\times$4$\times$4 = 256

Now, the number of ways in which the 4-digit numbers greater than 4321 can be formed is as follows:
Suppose, the thousand's digit is 4 and hundred's digit is either 3 or 4.
∴ Number of ways = 2$\times$4$\times$4 = 32
But 4311, 4312, 4313, 4314, 4321 (i.e. 5 numbers) are less than or equal to 4321.
∴ Remaining number of ways = $256-\left(32-5\right)=229$

Case II: Three-digit number
The hundred's digit can be filled in 4 ways.
Similarly, the ten's digit and the unit's digit can also be filled in 4 ways each. This is because the repetition of digits is allowed.
∴ Total number of three-digit number = 4$\times$4$\times$4 = 64

Case III: Two-digit number
The ten's digit and the unit's digit can be filled in 4 ways each. This is because the repetition of digits is allowed.
∴ Total number of two digit numbers = 4$\times$4 = 16

Case IV: One-digit number
Single digit number can only be four.
∴ Required numbers = 229 + 64 + 16 +4 = 313