3
$\begingroup$
  1. there are $9 * 9 * 8$ three digit numbers where all digits are different so $1000 - 9 * 9 * 8$ which have at least two digits that are the same and then there are $9$ numbers which have exactly three same digits so the number of three digit numbers where exactly two digits are the same is $1000 - 9 * 9 * 8 - 9 = 343$.

  2. Let's pick a digit between $1-9$. If we decide a three digit number should contain this digit twice then there are three positions for the remaining digit, when it's the first digit can be 8 kinds (144, 244, 344, 544 etc), the second and the three $9-9$ each (404, 414 etc) so far we have $9 * ( 8 + 9 + 9 )$ . Finally if the repeating digit is $0$ then the first digit can be any of $1-9$ so the end result is $9 * ( 9 + 9 + 9) = 243$.

Both can't be right. So which one is wrong and why?

$\endgroup$
  • 2
    $\begingroup$ There are $900$ three digit numbers, not $1000$, so there are $900-9\cdot 9\cdot 8$ three digit numbers that have at least two equal digits, not $1000-9\cdot 9\cdot 8$. $\endgroup$ – user236182 May 26 '16 at 17:53
1
$\begingroup$

In your first point, there are $900$ three digit numbers, not $1000$. This gives the missing $100$.

$\endgroup$

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.