2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Arthur Apr 9 '17 at 7:30
  • $\begingroup$ yeah i thought so. problem is case in which thousandth place is 4 also furthure splits up $\endgroup$ – Taylor Ted Apr 9 '17 at 7:32
  • $\begingroup$ That it does. But I still think that's the fastest way to go. $\endgroup$ – Arthur Apr 9 '17 at 7:32
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Barry Cipra Apr 9 '17 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.