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?
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asked Feb 28 '16 at 17:36
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12$\begingroup$ this is middle school probability question... $\endgroup$ – Oray Feb 28 '16 at 18:18
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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).
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$\begingroup$ 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
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$\begingroup$ @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
