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

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

• The thousands digit being $1,2$ or $3$ can be the same case (there are $3\cdot4\cdot4\cdot4$ such four digit numbers in total). However, you need to treat the thousands digit being $4$ specifically. – Arthur Apr 9 '17 at 7:30
• yeah i thought so. problem is case in which thousandth place is 4 also furthure splits up – Taylor Ted Apr 9 '17 at 7:32
• That it does. But I still think that's the fastest way to go. – Arthur Apr 9 '17 at 7:32

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.

• Note, this only counts the four-digit numbers meeting the OP's requirement. It might help to put it together with what the OP knows about counting the two- and three-digit cases (plus the one-digit case). – Barry Cipra Apr 9 '17 at 8:40