Assuming that letters can be duplicated there are 26 options for the 1st letter 26 for the 2nd and 26 for the third. You multiply these numbers together 26x26x26=17576 possible combinations.

Not to throw a fork in the gears but if for some reason you cannot duplicate letters there would be 26 options for the first, 25 for the second, and 24 for the third giving you 26x25x24=15600 options.

Now to for possibilities of his initials in any order you can list the possibilities EAM, EMA, AEM, AME, MEA, MAE giving you six or you can multiply again. For the first letter there are 3 options (A, E, or M) for the second there are 2 options, and for the third there is only one. So 3x2x1=6 So the probability of him getting a combination of his initals in any order is 6 out of 17576 (or 6 our of 15600 if letters cannot be duplicated)

To get his initial in the correct order there is only one option 1x1x1=1 making the probability 1 out of 17576. (or 1 out of 15600 if letters cannot be duplicated)

Hope this helps!

so for each letter location there is a 3/26 chance for the first, then 2/26 (because 1st letter would be used in the first spot) and then 1/26 because you need to make sure that this is the last of the 3 letters

3/26 x 2/26 x 1/26 Multiply then reduce 6/17,576 or 3/8788

For the probability in the exact order. You can only have that one letter as a choice so it is

1/26 x 1/26 x 1/26 Multiply 1/17,576

and 3/78