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Some 5 digit number can be formed without repeating the number 0 nevar there​

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I’m going to give this a shot and probably miss by a mile, but, oh, well. Hope it doesn’t mess up your homework too bad.

First, let’s figure out how many 5 digit numbers we can make with our 10 standard digits with no repeats.

So, how many of the digits could be used in the first place (left-most). That should be 9, because standard numbers don’t start with a zero.

The second digit place from the left could also be filled with 9 different numbers because we’ve already used one of the digits and there are no restrictions on what value can be second except that it can’t be the same as the first.

Going toward the left, we would be pulling from a set with 1 fewer members each time. So the 3rd, 4th, and 5th digits could be filled with 8, 7 and 6 different digits respectively. In all, we could form 9 x 9 x 8 x 7 x 6 different numbers from the beginning group. This gives us 27,216 numbers.

So, the last part of the puzzle is: How many of these are even numbers? My gut feel here is that exactly half of those numbers are going to be even because half of the digits are even and half are odd, so I’m predicting a grand total of 13,608 even numbers.

It will be interesting to see where I’ve gone wrong, here.

Plz mark my answer as branily and fol low me

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