probablity

mutually exclusive: union: add: a "or" b (since their sum must total to 1)
remember, that when not mutually exclusive a union b = pA + pB - p(A
intersection B). "The probability of A or B occuring"

intersection: multiply: a "and" b; same sample space. "The probability of both
a and b occurring"

counting

combinations: grouping matters but not the order in the group: n!/r!(n-r)!

combinations will be smaller than permutations. why? because we know just
that we have {a,b,c,d} in the set and we don't care about if its' {a,b,c,d} or
{a,d,c,d} etc.

Permutations and Combinations of Sets Not All Different: n!/n1!n2!n3!
(fOR example, we want combinations of different size subsets)
e.g.: "in how many ways can 7 scientists be assigned to one triple and two
double hotel rooms?"

Permutations is where we want a subset and we are looking for how many ways we
can combine those in the group we extract == n!/(n-r)! where r = the size of
our group to extract

The above will then tell us how many different sets we can have.

numbers of combinations of any size(in a circle): 2ⁿ -1

fundamental theory of counting:
P(n,r) = C(n,r) * P(r,r)

where C(n,r) = n!/r!(n-r)!
and P(r,r) = r!

so, that is because we need to take the permutations of the elements found in
the sets.