Hello cyt91 Originally Posted by

**cyt91** Calculate the number of even numbers between 5000 and 7000 which can be formed from the digits 1,5,6 and 8 if repetitions are allowed.

I tried working on this question a few times and came to 48 but, the answer given is 60. Can anyone help me??? Thank you!

I don't get either of these answers; I reckon it's 64. As follows:

The first digit is 5 or 6. So there are $\displaystyle 2$ choices here.

The second and third digits can be any one of $\displaystyle 4$; so that's $\displaystyle 4^2$ choices.

The final digit must be 6 or 8: $\displaystyle 2$ choices.

So the total number of choices, and hence the total number of numbers is:

$\displaystyle 2\times4^2\times2=64$

Grandad