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How many even three-digit numbers have distinct digits and have no digit $5$?

The answer my teacher gave was $252$, but I don't see how she got that. I thought it would be $6\times 8 \times 5=240$ because the $3\text{rd}$ digit must be even $(0,2,4,6,8)$, the $2\text{nd}$ digit can't be $5$ or the $3\text{rd}$ digit $(10-2=8$ options) and the first digit can't be $0$, $5$, the $2\text{nd}$ digit, or the $3\text{rd}$ digit ($10-4=6$ options). Any way I look at it, the last digit has to be even and there are $5$ options for even digits, so the final answer must end in a $5$ or a $0$, not a $2$. Please help!

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    $\begingroup$ You have to add those cases where $0$ is used up as the third digit and can't be there at the first digit. It would be better to split into 2 cases, one with last digit as $0$ and other with last digit as non-zero. $\endgroup$
    – gemspark
    Aug 24, 2020 at 4:11
  • $\begingroup$ Looks like they're counting 024 as a three digit number. $\endgroup$
    – David
    Aug 24, 2020 at 4:13
  • $\begingroup$ @ L_M Quite... while your logic is sound for the ones-place digit and the tens-place digit, your hundreds-place digit might have $6$ options (in the event the ones-digit was not zero) or it might have had $7$ options (in the event the ones-digit was zero). Your mistake is that you forgot to consider what happens if the list "zero, 5, 2nd digit, 3rd digit" had some overlap. $\endgroup$
    – JMoravitz
    Aug 24, 2020 at 4:14

3 Answers 3

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No. ending with non-zero digit = $4 .7.7=196$.

No. ending with zero digit = $1 .8.7=56$

P.S. The product above is (third digit options) x (first digit) x (second digit)

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I figured out a different way to do it actually. I see that you have to account for when the third digit is zero and non-zero because it changes the criteria for the first digit, but I considered this for the second digit as well. What I did was split it into three cases:

Last digit is zero: $7 \times 8 \times 1=56$ (second digit cannot be $5$ or $0$ : the first digit cannot be $5$, the second digit or the third digit)

Last digit is nonzero and second digit is zero: $7 \times 1 \times 4=28$ (first digit cannot be $5$, the second digit or the third digit)

Second and last digits are nonzero: $6 \times 7 \times 4=168$ (the second digit cannot be $5, 0$, or the last digit: the first digit cannot be $5,0$, the second digit or the third digit)

Addition principle:$56+28+168=252$

It gives the right answer, but is this reasoning sound?

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  • $\begingroup$ Looks sensible. $\endgroup$ Aug 24, 2020 at 6:39
  • $\begingroup$ You have just subdivided the non-zero last digit case of my answer into 2 subcases. $\endgroup$
    – gemspark
    Aug 24, 2020 at 7:27
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So, one is asking for the number of injections $f:\{1,2,3\} \to \{0,1,2,3,4,6,7,8,9\}$ where $f(1) \neq 0$ and $f(3) \in \{0,2,4,6,8\}$.

Ignoring the last two conditions, one gets $_{9}P_{3}=504$ possible numbers (possibly odd and/or having a leading zero).

If one requires the first digit to be nonzero, then one would then need to subtract the number of injections $\{2,3\} \to \{1,2,3,4,6,7,8,9\}$ to get $504-{_{8}P_{2}}=504-56=448$.

Odd numbers still have not been eliminated yet. We have to eliminate the cases where the last digit is $1, 3, 7,$ or $9$.

Suppose that the last two digits ($f(2)$ and $f(3)$) are $0$ and $1$ respectively. Then, one must have $f(1) \in \{2,3,4,6,7,8,9\}$, eliminating $7$ odd numbers to reduce the count to $441$.

Similar considerations apply when the middle digit is still $0$, but the last digit is now $3, 7,$ or $9$. This eliminates $3 \cdot 7=21$ more odd numbers, reducing the count to $420$.

Now, let's move on to the case where the middle digit is nonzero. Fixing a last digit $d \in \{1,3,7,9\}$, one would then need to subtract the number of injections $\{1,2\} \to \{1,2,3,4,6,7,8,9\} \setminus \{d\}$.

Suppose that $d=1$. Then, subtracting the number of injections $\{1,2\} \to \{2,3,4,6,7,8,9\}$ gives $420-{_{7}P_{2}}=420-42=378$. Forty-two odd numbers with the last digit equal to $1$ have now been eliminated.

Similar considerations apply when $d$ is $3,7,$ or $9$. This eliminates $3 \cdot 42=126$ more odd numbers, giving a final count of $252$.

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