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How many even numbers of four distinct digits greater than 5000 are possible? Please help me

The only thousand digit that are possible 5,6,7,8,9. The only hundred digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 The only ten digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 The only unit digit that are possible are 2, 4, 6, 8 5x9x9x4=1620

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  • $\begingroup$ You must have tried something. Where are you getting stuck? $\endgroup$ – lulu Dec 18 '17 at 21:04
  • $\begingroup$ If you have no idea at all, try to program it ! $\endgroup$ – Jean Marie Dec 18 '17 at 21:07
  • $\begingroup$ Of course I have an idea. The only thousand digit that are possible 5,6,7,8,9. The only hundred digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 The only ten digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 The only unit digit that are possible are 2, 4, 6, 8 5x9x9x4=1620 $\endgroup$ – SELİN KIRIŞ Dec 18 '17 at 21:10
  • $\begingroup$ Your calculation badly overcounts because the same digits might appear in each slot. For instance, you allow $6$ in every slot so your method counts $6666$ as a possible case which it is not. You have to ensure that you don't use the same digit multiple times. $\endgroup$ – lulu Dec 18 '17 at 21:19
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    $\begingroup$ It is your problem, not mine. You said "distinct integers". To me, that means the digits have to all be different. If you meant something else by it, please explain. $\endgroup$ – lulu Dec 18 '17 at 21:41
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There might be a prettier solution to this problem, but I would use some variation of a decision tree:

enter image description here

The layers being the digits chosen left to right starting with the thousands, and the beige and blue circles being the odd and even options respectively. Each branch has the product taken and the different (exclusive) alternatives summed to 1288.


edit

fleablood's answer gives a far better order of choosing the digits: first, then last, then others. A diagram for this might be:

enter image description here

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    $\begingroup$ Well, on the one hand that is certainly the most difficult way to do it correctly and it'd be a far smaller picture if the one's and thousands were the first two rows so the third and fourth row would have no branching. But on the other hand this is a GREAT visual that shows how to think logically and consistently and how to handle any minor case snags that come up. $\endgroup$ – fleablood Dec 19 '17 at 17:31
  • $\begingroup$ @fleablood Good point about the order of choosing. $\endgroup$ – Joffan Dec 19 '17 at 17:48
  • $\begingroup$ Nice answer with very nice graphics. Can I ask you which software you use for your figures ? $\endgroup$ – Jean Marie Dec 19 '17 at 18:56
  • $\begingroup$ @JeanMarie it's Powerpoint 2016 $\endgroup$ – Joffan Dec 19 '17 at 19:05
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THe first digit can be $5,6,7,8,9$. Those are four posibilities.

The second digit can be any of the ten $0,1,2..., 9$ (for some inexplicable reason you didn't include $0$) but the second digit must be different from the first. So there are $9$ options.

The third digit must be different from the first two so there are $8$ options.

The fourth must be different for the first three so so there are $7$ options.

The last digit must be even so it is $0,2,4,6,8$ and it must be different that the first four and ... we have no idea how many of the first four are even or not. So we are found in the Alps. Dang.

Start over.

Do two cases. Either the first number is even $6,8$ (2 options) or it is $5,7,9$ (3 options).

That last digit must be even so if the first is even then the second must be different so there are $4$ options because it must be different. If the first is not even there are $5$ options.

The second digit is different that the first or last so there are $8$ options.

The third digit must be different than the other three so there are $7$ options.

So if the first digit is even there are $2*4*8*7$. And if the first digit is not even there are $3*5*8*7$.

So there are $2*4*8*7 + 3*5*8*7 = (2*4 + 3*5)*8*7 = (8+15)*56 = 23*56= 1288$ such numbers.

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  • $\begingroup$ I suggest deleting everything above "Do two cases" except the observation that the digit $0$ can be used since you seemed to be solving a different problem involving five digits rather than the problem at hand. $\endgroup$ – N. F. Taussig Dec 19 '17 at 11:09
  • $\begingroup$ I was trying to go through the thought process of how to figure out the strategy in the first place. To realize one must do the restricted cases first rather than in arbitrary order and to realize that conditional cases are needed are a painful but necessary part of the learning process. $\endgroup$ – fleablood Dec 19 '17 at 17:25
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The question says distinct digits, so if one digit appears in the hundreds place, it should not appear again in the tens place as example.

Take the case for 5 in the thousands place. We have five choices for the ones place $\{ 0,2,4,6,8 \}$. If we choose any digit for the ones place, that digit will be prohibited in the tens and hundreds places.

This means that for the hundreds place two digits are prohibited (5 for the thousands place and a digit for the ones place), so we have 8 choices for the hundreds place.

Similarly, now for the tens place three digits are prohibited (5 for the thousands place, a digit for the ones place, and another for the hundreds place), so we have 7 choices for the tens place.

So, we have 1 * 5 * 8 * 7 choices for even numbers of four distinct digits greater than 5000 (that starts with 5).

You can consider the cases where the thousands place takes 6, 7, 8, or 9.

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  • $\begingroup$ "two digits are prohibited so we have 9 choices" I was under the impression that $10 - 2 \ne 9$. $\endgroup$ – fleablood Dec 18 '17 at 21:48
  • $\begingroup$ thanks for the hint, I edited the answer based on your feedback. $\endgroup$ – Student Dec 18 '17 at 21:52

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