How many 3-letter words without repeating them can be made up of alphabet {a, b, c, d, e, f} in which the letter e or the letter f or both are used? How many 3-letter words without repeating them can be made up of
alphabet {a, b, c, d, e, f} in which the letter e or the letter f or both are used? 
with permutations
 A: So, you tried to calculate it by seeing how many have only "e" or only "f" and then both. This can get rather complicated so let us take a different approach:
The number of 3-letter words where "e" is the first letter is $5*4$ possibilities. This is identically true for "f" as the first letter. Now, let's count the number of 3-letter words where "e" is the second letter but "f" is not the first letter (so as not to double count): this is $4*4$ possibilities. Identically so for the case where "f" is the second letter but "e" is not the first letter. Finally, let's count the number of 3-letter words where "e" is the third letter but "f" is not the first or second letter(so as not to double count):  this is $4*3$ possibilities. Identically so for the case where "f" is the third letter but "e" is not the first or second letter. So, in total, we have $20*2+16*2*12*2 = 96$ total possible 3-letter words that satisfy the condition.
Alternatively, we can count the number of 3-letter words that don't have either "e" or "f" and subtract that number from the total number of possible 3-letter words creatable using the alphabet: $6*5*4-4*3*2 = 96$ total possible 3-letter words that satisfy the condition.
If you would like me to clarify anything please let me know :)
A: Divide given letters $a, b, c, d, e, f$ into two groups $\{a, b, c, d\}$ & $\{e,f\}$
Now, select any two letters out of $\{a, b, c, d\}$ by $^4C_2$ different ways  & any one letter out of $\{e,f\}$ by $^2C_1$ different ways & arrange these three selected letters by $3!$ ways to form 3-letter words having either $e$ or $f$. Therefore, the total number of 3-letter words having either letter $e$ or $f$ $$^4C_2\times ^2 C_1\times 3!=72$$
Similarly, select any one letter out of $\{a, b, c, d\}$ by $^4C_1$ different ways  & two letters out of $\{e,f\}$ by $^2C_2$ different ways & arrange these three selected letters by $3!$ ways to form 3-letter words having both $e$ & $f$. Therefore, the total number of 3-letter words having both letters $e$ and $f$ $$^4C_1\times ^2 C_2\times 3!=24$$
Hence the total number of 3-letter words required
$$72+24=96$$
A: The total number of three lettered words is 
6C3(choosing any three letters out of 6)*3!(arranging them)
Number of words where e and f are both NOT used is 
4C3*3! (Same logic as above)
The required answer is “Total number of words-number of words not containing e or f”
That is 6C3*3!-4C3*3! = 96
