How many $4$-digit even numbers have all $4$ digits distinct
If the number ends with a $0$ then there are $9$ choices for the first digit, $8$ for the second and $7$ for the third, which makes $1\times9\times8\times7= 504$ possibilities.
If the number is even ending with something else than $0$ then there are $4$ choices for the last digit, $8$ choices for the first digit (no $0$ nor the last digit), $8$ for the second digit and $7$ for the third digit, which makes $4\times8\times8\times7 =1792$
Together, this gives $2296$ numbers with $4$ distinct digits that are even. Note that this does not allow leading $0$, as you see to want it based from the question.
Suppose if count all the 4 digit numbers not starting with 0, I would have 9 x 9 x 8 x 7 = 4536 (9 choices for first digit excluding 0, 9 choices for the second digit excluding the first chosen digit, followed by 8 and 7 for the other two digits)
Out of 4536 half should be odd and the other half should be even, so 4536/2 = 2268 odd and 2268 even numbers
But my answer is not matching with the above solution, where have I gone wrong ?
I could not understand what you meant by Now, this makes the selected digit not being among the unit digit. Can you please tell some numbers which are missed out ?
"Out of 4536 half should be odd and the other half should be even, so 4536/2 = 2268 odd and 2268 even numbers"
How did u conclude that there are equal amount of odd and even numbers? Those 4536 numbers are not consecutive. They are numbers having DISTINCT digits.
Puja Mishra yes it is wrong.The number can't start with 0.for example 0372 should not be accepted.
It may be your mistake :-
suppose at 4th position you selected 0
then for the first position you have 9 choice (1-9) and not 8 choices