How many even four digit numbers have no repeating digits? How many even four digit numbers have no repeating digits?
I got $5400$. Is this correct?
 A: If it ends in $0$ there are $9\cdot8\cdot7$ options, for first, second and third digit.
Otherwise there are $4$ options for the last digit, $8$ options for the first digit (it can't be $0$ or the same as the last digit), $8$ options for the second digit and $7$ options for the third digit.
So the answer is $9\cdot8\cdot7+8\cdot8\cdot7\cdot4=2296$.
A: I think $5400$ is a bit an overestimate .. an upper bound (counting only even numbers) is given by
$$\frac{9999 - 1000 + 1}{2} = 4500$$
Anyway consider counting with ending $0$ or not


*

*$_ _ _ 0$ $\Rightarrow$ you have $$9 \cdot 8 \cdot 7$$ choices as there are $9$ possible digits for the third digit, $8$ for the second and $7$ for the first

*$_ _ _ _$ $\Rightarrow$ the number must be even so you have $$4 \cdot 8 \cdot 8 \cdot 7$$ possibilities because there are $4$ choices for the last digit (the number must be even), $8$ choices for the first digit (the first cannot be $0$), $8$ choices for the second and $7$ for the first.
So, summing up the partial sums you get
$$9\cdot8\cdot7+4\cdot8\cdot8\cdot7=2296$$

A: Assuming that 9898 has no "repeating" digits:
The first digit cannot be 0, so there are nine choices, 4 even and 5 odd. 
For the second digit we have nine choices. If the first digit is even then we have four even and five odd choices, otherwise five even and 4 odd. Total is 81 choices, 41 even and 40 odd. 
For the third digit we have nine choices. Again four even and five odd choices if the second is even, five even and 4 odd otherwise. Total 729 choices, 364 even and 365 odd. 
The last digit must be even. So we have four choices if the third is even and five choices if the third is odd. Total 4x364 + 5x365 = 9x364 + 5 = 3281. 
