Yes that's correct. Another way to do it, would be to split it up into 2 disjoint cases, one where the last digit is 2, and one where it's 6, thus you would get 5*5*5*5*1+5*5*5*5*1 = 625+625 = 1250. But your version is prettier
Given numbers 2, 5, 6, 7, 9. Find the number of ways of arranging a five digit even number from the given numbers?
If we assume replacement then we would have 5 choices for the 1-4 digit and then 2 choices (2 and 6) for the 5th digit. So, by the counting principle,
5*5*5*5*2=1250.
Is this correct?