discrete mathematics - Counting with Combinatorics (Just need my work checked over, pretty sure it's right)

Consider the alphabet A,B,C,D,E,F and make words without repetition of letters allowed.

A.) How many six-letter words are there?

B.) How many words begin with D or E?

C.) How many words end in B or A?

D.) How many words begin with D or E and end in B or A?

E.) How many have first letter neither D nor E and last letter neither B nor A?

A. The total number of six-letter words without repetition of letters:
6 letters. = 6! = 720

B. How many words begin with D or E? -- Seeing that we're using 1 of the 6 letters to start off with, the answer will remain the same for both. Beginning with D would remove the first letter from the equation. Therefore, the answer would be 5!. Same for beginning with E. Correct me if I'm wrong! Do I add these results together (5! + 5!=240)?

C. How many words end with B or A? --Same concept as part B, right? We take the last letter out of the six-letter word equation and we're left with 5. =5!+5!=240 for both ending in B or A.

D. Here's where I may get confused. How many words begin with D or E and ends with B or A? -- There would be 4 cases? Begins with D and ends with B, begins with D and ends with A, begins with E and ends with B, begins with E and ends with A. Each case would retrieve the same result? Therefore, we know what the first and last letters are. We just need to calculate the middle 4 letters that are not repeated. That equals 4!. 4 cases, so 4!+4!+4!+4!=96. Once again, correct me if I'm wrong.

E. How many have first letter neither D nor E and last letter neither B nor A? --This means the first letter can be A,B,C, or F. This gives us 4 total choices. The last letter can be C,D,E, or F. This also gives us 4 total choices, but we can't repeat letters. C and F are the common letters. So, to calculate the first letter combinations, we can do 4!/((4-2)!)=12 combinations. Same with the last letter. And, for the middle 4 letters that can't be repeated, we have 4! combinations. Therefore to solve for the total combinations, it would be 12+4!+12= 48 combinations.

Just looking for someone to check over my work! I'm not 100% sure that I did this all correctly. Thanks!

Answer

E) Call such words good. We will count the bad words and subtract from $720$.

There are $240$ bad words that begin with D or E, and $240$ that end with A or B. Add. We get $480$. However this double counts the $96$ bad words that begin with D or E and end with A or B. Thus the number of bad words is $240+240-96$.

Now we can find the number of good words.

Remark: The solutions to A, B, C, D are correct.