# Math Help - permutation

1. ## permutation

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?

2. 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

3. Originally Posted by sfspitfire23
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?
With that assumption, that answer is correct.
However, I would argue that the phrase, “the number of ways of arranging a five digit even number from the given numbers” implies non-replacement.