How many natural numbers not exceeding $4321$ can be formed using digits $4,3,2,1$ if digits can repeat?

Question

How many natural numbers not exceeding $4321$ can be formed using digits $4,3,2,1$ if digits can repeat ?

now first i consider two digit numbers which can be 16. Now 3 digit numbers which are 64. my problem is to find 4 digit numbers. should i make 4 cases and add them all up in each case thousandth place being 1,2,3,4 respectively?

thanks

Answer 1

Using these four digits we have $4^4=256$ different numbers as a total.

However, some of the form $4XYZ$ would exceed $4321$.

Among the latter numbers there are $4^2=16$ of the form $44XY$.

Regarding those of the form $43XY$, if $3,4$ are used for a third digit then the resulting number would exceed $4321.$ There are $2\cdot 4=8$ such numbers.

Then among those of the form $432X$ $3$ would exceed $4321$. So, we have

$$256-16-8-3=229$$

appropriate combinations.