I am super confused with permutations and combinations. I just don't see when you are supposed to use each one, and when you aren't supposed to use either.

It would really help me if someone could check these. Please?

In a very general explanation,
in a permutation the order or the position of your arrangement matters,
while in a combination the order does not matter.

e.g. Suppose from a class of 25 students, 6 are to be chosen to form a mathteam
The order in which they are chosen does not matter, so that would be a combination.

On the other hand if those 6 students are to become experts in different areas of math on the team, that would be a permutation.

1. This is one of those special cases.
I will asssume that there is no specific order or special chairs to be filled.
So let's fill the first chair, that would be 10 ways, the second in 9 ways, etc.
so that would be 10! ways, right?
Now suppose everybody got up and moved to the chair on their left, and then sat down. Would the arrangement of people sitting around the table have changed? NO!
Now think of how many ways they could do this? Could they not do this 10 times, each time resulting in the original arrangement?
So the 10! ways has to be divided by 10, giving you 9! ways.

3. for the license plate digits may not be repeated, but letters may.
Can the plate start with a 0? I will assume it can.
so the first 4 numbers are 10*9*8*7
followed by 2 letters which would be 26*26

so the number of plates is 10*9*8*7*26*26

4. you have 9 different letters, of those arrange 6, (the order matters)

a) the number of ways would be 9*8*7*6*5*4

b) they can be repeated, so 9*9*9*9*9*9
or 9^6

c) 3 ways to fill the first spot with a vowel, after that anything goes without repeating the letters

so no of ways = 3*8*7*6*5*4

d) only change is 3*9*9*9*9*9

Let me think about #2, it seems like it should be a probability question
which would be P(9,6)

the very last line of "which would be P(9,6)"
should have come after 4.a) to read

" a) the number of ways would be 9*8*7*6*5*4
which would be P(9,6)"

2. I am a bit mystified by the wording.
They are using the word "selected" which suggests a combination, in other words, the order does not matter
in a) there is only one way for 4 bags to be red, namely RRRR

b) by the same thinking there is only one way for 4 bags to be 2 blue and 2 red,
BBRR (BRRB would be the same since order does not matter)

c) exactly 3 of the 4 bags must be red, meaning that the 4th could be either blue or green
which would be 2 ways: RRRB or RRRG

Oh ok, I get it. But i have one more question.

Why are there always more ways when you use a permutation? Wouldn't you think that you would have more ways when you used a combination since the order doesn't actually matter? I don't know if that makes sense to you. It's kind of bugging me

suppose from you 5 friends you "choose" 3 to ride in your car.
That would be C(5,3) = 10

now suppose you pick 3 people to sit in the passenger seat, the rightrear seat and the leftrearseat.
that would be 5*4*3 or P(5,3) = 60 ways

notice the permutation is more.

Another way to look at it:
suppose you pick one of the 10 combinations from my example.
once you have selected the 3 people you want you could now fill the
frontseat, rearleft and rearright seats in 3*2*1 or 6 ways
And of course we could do these 6 arrangements for each of our 10 choices, giving us a total of 60 arrangements.

(I am still puzzled by #2. Are you studying probability questions yet?)

(fog)(-2)