MHB Finding Even Natural Numbers w/ No Repetition: 0-5 & 6

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The discussion focuses on calculating the number of even natural numbers less than 100,000 that can be formed using the digits 0, 1, 2, 3, 4, 5, and 6 without repetition. It establishes that the last digit must be even, leading to four possible choices: 0, 2, 4, or 6. The calculations break down the total into one-digit, two-digit, three-digit, four-digit, and five-digit numbers, yielding 3, 21, 90, 300, and 600 even numbers respectively. The final total sums to 1,014 even natural numbers that can be formed under these conditions. This methodical approach ensures that all criteria for forming valid numbers are met.
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The number of even natural numbers less than 100000 that can be formed from the digits of the set (0,1,2,3,4,5,6) so that the digits in the number are not repeated is?
Here I understand that the even number in the last place is an even number, that is, it has 4 possibilities, but won't the numbers repeat themselves?
 
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As you say, a number is even if and only its "ones" place is even as you say.
I would count the number of one-digit, two digit, three digit, four digit, and five digit numbers.
(Do you see why I do not include six and seven digit numbers?)

There are three such numbers with one digit: 2, 4, and 6. (0 is not a natural number.)

For two digit numbers, once we have one of 2, 4, and 6 as the ones digit, there are five possible 10's digits (again, 0 cannot be the 10's digit in a two digit number.) That gives 3(5)= 15 such numbers. If 0 is the one's digit, then there are 6 possible digits for the 10's digit so that is another 6 giving 6+ 15= 21 two digit even numbers,

For three digit numbers, we can have anyone of those 15 two digit numbers that do NOT have a 0 with the remaining four non-zero digits so 4(15)= 60. Of the six that do have a 0, we can take any of the remaining 5 digits as the new digit, The are 5(6)= 30 such numbers so 60+ 30= 90 such three digit numbers.

Similarly, of the 60 three digit numbers that do NOT have a 0 we can add any of the remaining three non-zero digits so 60(3)= 180. Of the 30 numbers that do have a 0 we can add any of the remaining four non-zero digits. There are 30(4)= 120 such numbers so 180+ 120= 300 such four digit numbers.

Finally, of the 180 four digit numbers that do NOT have a 0 we can add any or the remaining 2 non-zero digits so (180)(2)= 360. Of the 120 four digit numbers we also add only the two non-zero digits because the first digit in a number cannot be 0, That is (120)(2)= 240 so there are 360+ 240= 600 such five digit numbers.

That gives a total of 3+ 21+ 90+ 300+ 600= 1014 even numbers, under 100000, using those digits at most once.
 
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