Find the probability of rolling ten different, standard 6-sided dice simultaneously and obtaining a sum of 30? Find the probability of rolling ten different, standard 6-sided dice simultaneously and obtaining a sum of 30?
I started to answer this question by setting up an equation like this:
x1+x2+...+x10=30
with 0 less than or equal to xi less than or equal to 6
But we know the die will have values 1-6 so we have
y=x+1 so that
y1+y2+...+y10=20
Now I know we can have at most 4 6's, 5 5's, 6 4's, 10 3's, 10 2's, 10 1's.
I was not sure if I am heading in the right direction or should I just find all possible outcomes of the die and use the inclusion-exclusion method to find all the permutations in which the sum will be 30?

Say that $n$ represents the number of dices, $x$ the total sum and $f(n, x)$ the total number of different ways in which this sum can be obtained.
We have the following recurrence relation:
$$f(n,x)=\sum_{i=1}^6 f(n-1, x-i)\tag{1}$$
...which baciscally says that you can calculate $f$ by assuming that the first dice can roll 1,2,3,4,5 or 6 and by adding the number of ways in which you can roll the sums $x-1,x-2,\dots x-6$ by using one dice less.
We have the following exit conditions:
$$x \lt 0 \implies f(n,x)=0\tag{2}$$
$$f(0,0)=1\tag{3}$$
In this particular problem you want to calculate $f(10, 30)$ and then divide the result with the total number of different rolls (which is $6^{10}$). You can complete the task with this miniscule Python script:
cache = dict()

def f(n, x):
    if x < 0: return 0
    if x == 0 and n == 0: return 1
    key = (n, x)
    if key in cache: return cache[key]
    cache[key] = sum([f(n - 1, x - i) for i in range (1, 7)])     
    return cache[key]

count = f(10, 30)
prob = count / (6 ** 10)
print("Number of ways to roll 30 with 10 dices:", count)
print("Probability:", prob)

And the result is:
Number of ways to roll 30 with 10 dices: 2930455
Probability: 0.048464367913724195

