5
$\begingroup$

A gang of thieves has discovered that the 5 digit combination to a safe is even and has exactly one odd digit and exactly two digits equal to each other. How many possible combinations are there for them to try?

|improve this question
$\endgroup$
  • 12
    $\begingroup$ this is middle school probability question... $\endgroup$ – Oray Feb 28 '16 at 18:18
8
$\begingroup$

The answer is:

$7200$

Argument:

  • There are four possible parity patterns for the combination (o=odd, e=even): $oeeee$, $eoeee$, $eeoee$, and $eeeoe$.
  • Every parity pattern contains four even digits, of which two must be equal to each other. This gives ${4\choose2}=6$ possibilities for picking the two equal positions.
  • Once we have fixed the parity pattern and the two positions with equal digits, there remain five possibilities for the odd digit ($1,3,5,7,9$) and $5\cdot4\cdot3=60$ possibilities for fixing the three distinct even digits ($5$ possibilities for the leftmost even digit, $4$ for the next, and $3$ for the last.)

Altogether, this yields $4\cdot 6\cdot 5\cdot 60=7200$ possible combinations (out of $100,000$ combinations overall with five digits).

|improve this answer
$\endgroup$
  • $\begingroup$ You're hiding something shifty when you say "5 possibilities for the leftmost even digit." I'm quite certain you're mistaken; the way you're counting the "two equal positions" vs. the way you pick "three distinct even numbers" is miscounting something. I'm having trouble putting my finger on it though. $\endgroup$ – Wildcard Mar 5 '16 at 2:32
  • $\begingroup$ @Wildcard Remember to fix the parity pattern and the position of the equal digits before doing it. There is no miscounting. $\endgroup$ – Fimpellizieri Mar 5 '16 at 3:22

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.