Monograms

First Solution : The possible monograms are $ABZ, ACZ, ..., WXZ, WYZ, XYZ.$ Any two-element subset of the first 25 letters of alphabet, when used in alphabetical order, will produce a suitable monogram when combined with $Z$ . For example, the subset ${L,J}={J,L}$ will produce $JLZ$ . Furthermore, to every suitable monogram there corresponds exactly one two-element subsets that can be formed from a set of 25 letters, and there are $\binom{25}{2}=300$ such subsets.

Second Solution : When Z is fixed as the last initial. If the first initial is A, there are 24 choices for the middle initial, e.g., $B,C,D,...,Y$ . Similarly, there are 23 choices for the middle initial if the first initial is B, e.g. $C,D,E,...,Y$ . Continuing in this way there are a total of $24+23+22+...+1=\frac{24(25)}{2}=300$ suitable monograms.