Number of strings that start and end with the same letter or are palindromes or contain vowels only 
In this question, we consider strings consisting of 26 characters, with each character being a uppercase letter. Determine the number of such strings that 
  (a) start and end with the same letter, or 
  (b) are palindromes, or 
  (c) contain vowels only

For this problem, I approached each scenario a, b, c separately. I then joined the 3 with an OR to satisfy the question in the final answer.
a. $26$ characters, if the last letter must be the same as the first, we really only have $25$ characters to choose from, so part a = $26^ {25}$
b. $26$ characters, but if it is a palindrome, the length is cut in half, giving us $26^{13}$. 
c. If the characters are vowels only, there are only A E I O U, $5$ characters to choose from. And we have a full 26 characters to pick. $26^{5}$
The three value then gets OR'd together
$26^ {25}$ OR $26^{13}$ OR $26^{5}$  
Is what I did correct? I'm working from practice problems and am not given an answer key so it's hard to really practice when I don't know if I'm doing it right or wrong. Thank you!
 A: The correct answer is
$$26^{25}+5^{26}-5^{25}$$
There are two key steps in getting the answer.  First, condition (b) is irrelevant: palindromes necessarily start and end with the same letter, so they've already been accounted for under condition (a).  Second is the inclusion-exclusion principle applied to conditions (a) and (c):  After adding together the counts for each condition separately, you subtract the count of their intersection, namely the number of vowel strings that begin and end with the same vowel.
The OP mistakenly got $26^5$ instead of $5^{26}$ for condition (c); $26^5$ counts the number of $5$-character strings, with each character being any letter of the alphabet. (When forming a string of vowels, there are $5$ choices for the first character, $5$ for the second, and so forth, for a total of $5\times5\times\cdots\times5$.)
A: (a) Let me consider that you pick up the first letter is same and the last letter is automatically the same. therefore  $26^{25}$
(b) I think this one is complicated to explain, you need to consider 1 same numbers pair and other 24 letters are the same, 2 pairs of the same numbers and different to other pairs and the rest of 22 letters are the same... 13 pairs of the same numbers and different from other pairs. therefore  $26P1 + 26P2 + 26P3  + 26P4 + 26P5 + 26P6 + 26P7 + 26P8 + 26P9 + 26P10 + 26P11 + 26P12 + 26P13$.
(c) You have only vowels in 26 letters of a string. So, you should consider only vowels for 26 letters in a row. Therefore,
$5^{26}$
