Consider the alphabet {a, b, c, d, e} Select at random a word consisting of four letters.
What is the probability that all four letters of the word are distinct?
What is the probability that the word contains no vowels?
What is the probability that the first letter of the word is a vowel?
Let X denote the number of consonants in a particular four-letter word. What is the expected value E(X)?
for a.) I got 120 just by doing 5*4*3*2 = 120
for b.) I got 2 different ways: either 3*3*3*3 and getting 81 or by getting just 3/5
for c.) i got 2/5, but i'm pretty sure my answer for b and c are wrong considering i didn't use a perumatation for them
and then d.) i have absolutely no idea how to do that andwould like help with that
The very first step to this series of questions is to try to decide on a convenient sample space to use.
Our sample space in this problem is the set of four letter words which use the letters $\{a,b,c,d,e\}$, and I will denote the set by the name $S$. This is a particularly useful sample space because the problem implies that we are selecting one of these words uniformly at random, so each outcome in the sample space is equally likely, i.e. it is an equiprobable sample space.
When we have an equiprobable sample space like this, the probability of an event occurring is the ratio of the number of outcomes in the event over the number of outcomes in the sample space. That is to say $$Pr(A)=\frac{|A|}{|S|}$$
(note: this is ONLY guaranteed to be true in an equiprobable sample space. if the outcomes are not equally likely the formula above might not be true)
In our question here, the size of $|S|$ can be calculated using multiplication principle and we learn that $|S|=5^4$
For part (a), you correctly calculated the number of words with all different letters to be $5\cdot 4\cdot 3\cdot 2$. Letting $A$ be the event that all letters are different, we have then $|A|=5\cdot 4\cdot 3\cdot 2$. This implies then that $$Pr(A)=\frac{|A|}{|S|}=\frac{5\cdot 4\cdot 3\cdot 2}{5^4}$$ as you correctly eventually got in the comments above.
For part (b), again you correctly calculated the number of words with all letters being consonants (i.e. no vowels) to be $3\cdot 3\cdot 3\cdot 3=3^4$. Letting $B$ be the event that there are no vowels, that is then to say that $|B|=3^4$
We see then that $Pr(B)=\frac{|B|}{|S|}=\dots$
For part (c) you could approach directly with counting to find that there are $2\cdot 5\cdot 5\cdot 5$ words which begin with a vowel to get $Pr(C)=\frac{|C|}{|S|}=\frac{2\cdot 5^3}{5^4}=\frac{2}{5}$, but it is probably easier to think about it directly using a probabilistic argument by noting that only the first letter matters in our calculation.
For part (d), as already mentioned in the comments above linearity of expectation implies that $$E[X+Y]=E[X]+E[Y]$$
Letting $X_i$ be the random variable $X_i=\begin{cases}1&\text{if the}~i\text{'th letter is a consonant}\\0&\text{otherwise}\end{cases}$ we see that the total number of consonants in our word is $X=X_1+X_2+X_3+X_4$
We have then that $E[X]=E[X_1]+E[X_2]+E[X_3]+E[X_4]$
Now, remember the definition of expected value. $E[X]=\sum\limits_{k\in \Delta} k\cdot Pr(X=k)$ where $\Delta$ is the set of possible values that $X$ can take.
In our specific case, $E[X_1]=0\cdot Pr(X_1=0)+1\cdot Pr(X_1=1)$ which simplifies as $E[X_1]=Pr(\text{the first letter is a consonant})$. Similarly each of the other terms can be simplified and using this knowledge one can calculate $E[X]$.
As a final aside, you said "I'm pretty sure my answers to b and c are wrong because I didn't use a permutation for them."
There are many questions in probability and combinatorics where permutations are not used, just like there are many questions where catalan numbers and stirling numbers aren't used. Permutations and combinations should be some of the tools in your toolbox that get used on occasion for problems that need it. As you continue your studies you will continue to grow your toolbox, increasing the variety of problems that you can solve, but don't rely too heavily on your toolbox as there will eventually be problems that are truly new to you that requires a technique you've never seen before.