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So what I’ve currently done is break it down into the following ends: 4_______2 4_______4

And etc, creating endings that would fit the question in asking for numbers that are both larger, and even. But I’m not sure how to proceed with the rest?

Is there a proper approach rather than guessing?

James Baxter
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  • Umm.... why not, say, $9________2$? – David G. Stork Apr 08 '21 at 20:54
  • I do not want to use guesswork, but that is already in my list of possible start and endings. – James Baxter Apr 08 '21 at 20:54
  • @JamesBaxter I don't think you understand David G. Stork's comment. – Jacky Chong Apr 08 '21 at 20:58
  • @JackyChong. You're right, he doesn't understand it. Thanks. – David G. Stork Apr 08 '21 at 20:58
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    Care to explain it then? – James Baxter Apr 08 '21 at 21:02
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    How is 4_______4 possible? – user Apr 08 '21 at 21:17
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    @JamesBaxter The point David G. Stork is making is that your initial list of ends is incomplete because an acceptable number also can start with $7$ or $9$. So it's not good enough to count just the combinations you've listed. Also, since you're only supplied $8$ digits and you need to end up with an eight-digit number, you can't repeat $4$. – Robert Shore Apr 08 '21 at 21:31
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    I would have liked to answer this question but I do not see your working other than a statement that you split into cases, without listing all cases and without showing your work. – Math Lover Apr 08 '21 at 21:31
  • Please make an effort to solve the problem yourself even if the method is not the most efficient. In any case, one approach would be to simply count all possible even numbers (which is straightforward) and then subtract all even numbers that start with $2$ or $3$ - again easy to count. – Math Lover Apr 08 '21 at 21:34
  • Hint: lead 9, lead 7, lead 4. $\binom{6}{3,2,1}\cdot2+\binom{6}{1,2,2,1}\cdot2+\binom{6}{1,3,2}$ – 2'5 9'2 Apr 08 '21 at 21:35

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