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

In summary, 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) without repeating any digits is 1014. This includes one-digit, two-digit, three-digit, four-digit, and five-digit numbers and excludes numbers with six or seven digits.
  • #1
stamenkovoca02
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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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  • #2
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.
 

1. How do you find even natural numbers with no repetition from 0-5 and 6?

To find even natural numbers with no repetition from 0-5 and 6, you can simply list out the numbers and eliminate any duplicates. In this case, the numbers would be 0, 2, 4, 6.

2. Why are we only looking at even numbers and not odd numbers?

In this specific scenario, we are only looking at even numbers because the instructions specify to find even numbers. However, if the instructions were to find all natural numbers with no repetition, then we would also include odd numbers.

3. Can we use a formula to find these numbers instead of listing them out?

Yes, there are formulas that can be used to find even natural numbers with no repetition. For example, the formula 2n can be used to generate even numbers starting from 0. However, in this case, it would still require listing out the numbers and eliminating duplicates.

4. How do you know if a number has already been used and should be eliminated?

In this scenario, we are looking at a small range of numbers, so it is easy to keep track of which numbers have already been used. However, if the range was larger, one way to keep track would be to use a list or array to store the numbers that have already been used and check against that list before including a number.

5. Can this method be applied to finding even natural numbers with no repetition in a larger range?

Yes, this method can be applied to finding even natural numbers with no repetition in a larger range. However, as the range gets larger, it may be more efficient to use a formula or a different method to find these numbers.

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