Permutations / Combinations - suppose a word is a string of 8 letters of the alphabet with repeated letters allowed 1.) How many words are there?
Not sure how to solve this since repeated letters are allowed.  $n^r$ is the formula we are told to use for permutations with repeated objects, but $26^8$ seems like too large of a number.
2.) How many words end with the letter T?
I assume this would be a combination of the form $26 \choose 1$
3.) How many words begin with R and end with T?
I assume this would be a combination of the form $26 \choose 2$
4.) How many words start with A or B?
I assume this would be a combination of the form $26 \choose 1$ $+$ $26 \choose 1$
5.) How many words begin with A or end with B?
Same as number 4?
 A: 1: $26^8$ is correct.  There are 26 choices for the first letter, 26 for the second letter, etc.  You multiply all the 26's because these choices are independent. "multiplication principle"
2: The last letter is known.  Otherwise, you choose the first 7 letters freely, so it's $26^7$.
3: I leave for you.
4: This is the "addition principle".  Count words beginning with A, and words beginning with B.  These form two disjoint sets, as a word cannot begin with both A and B.  Hence you add the two answers.
5: This is not the same as 4, because now it is possible to both begin with A and end with B.  Such words were counted twice.  Hence, you need to add the two answers, but now subtract the double-counted words.  How many double-counted words are there?  Such a word begins with A and ends with B, so there are 6 letters in the middle freely chosen.
A: *

*$26^8$ is correct. There are $26$ options for each of the $8$ letters. 

*$26^7$, since you can choose the first $7$ letters, but the last one has to be T.

*$26^6$, since you can only choose the middle $6$.

*You can choose the last $7$ letters. There are $2$ options for the first one. What do you get?

*Do words beginning with A, ending with B, and both beginning with A and ending with B separately. 
