Discrete Probability: Four dice are thrown, what's the probability that... Four dice are thrown, what's the probability that:

a) None of them fall higher than three?

b) None of them fall higher than four?

c) That four is the highest number thrown?

So for a, I wanna think about the denominator first. There are 6 possible outcomes for each dice, and we have four dice. So we're technically picking r out of n objects, aka picking 4 possible values out of 6. So the denominator/sample space should be $6^4$ right?

Now this is where I'm getting tripped up. Order technically shouldn't count, because we only care about the quadruplets that don't have a value higher than 3. I might be wrong, maybe I am, but could someone explain a bit more? My professor stated that typically in the sample space/denominator, we want order to count. 

So I'm trying to think of the numerator now, so we want to find the probability that no values appear higher than 3, so some events that can occur are: $(1, 1, 2, 3), (1, 2, 2, 3), (1, 1, 1, 1)$ etc. Again, I can't see why order should count here, because if we only are concerned about what appears rather than how they appear, then we can say $(1, 1, 2, 3)$ is equal to $(1, 2, 1, 3)$? So if order does not count, and we have replacement, then does this lead the case where we use $$\binom{n-1+r}{r}$$to find out the probability? 

I'm pretty sure if I can do a), I could prob do b), but if someone could maybe lead me in the right direction for c) I would appreciate that too!
a) The probability that a die does not fall higher than $3$ is given by
$$P(X\le3)=\frac{3}{6}=\frac{1}{2}$$
So as each event is independent we can find the probability that no die falls higher than three by raising this result to the power of $4$
$$(P(X\le3))^4=\frac{1}{2^4}=\frac{1}{16}$$

b) Similarly the probability that a die does not fall higher than $4$ is given by
$$P(X\le4)=\frac{4}{6}=\frac{2}{3}$$
So the final probability is
$$(P(X\le4))^4=\frac{2^4}{3^4}=\frac{16}{81}$$

c) If $4$ is the highest rolled value then every die rolled has a value $\le 4$. But we also need at least one $4$ to be rolled - so we need to subtract the probability of rolling every dice $\le3$. The answer is then
$$(P(X\le4))^4-(P(X\le3))^4=\frac{16}{81}-\frac{1}{16}=\frac{175}{1296}$$
which is the difference of the answers from a) and b).
$a)$
The probability that one dice rolls a number equal to $3$ or lower is $\frac{3}{6}=\frac{1}{2}$.
Hence, the probability that all of them roll $3$ or lower is $(\frac{1}{2})^4=\frac{1}{16}$.

Can you do the same for $b)$?

$c)$
Firstly, There is a probability of $(\frac{4}{6})^4=(\frac{256}{1296})$ that all numbers rolled are $4$ or lower.

However, at least one of the numbers must equal $4$, so we subtract the probability that they are all 3 or lower. We already calculaties this in $a)$.

Hence, the probability that the highest number rolled is $4$, equals
$$\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$