Can someone please clarify combinations vs permutations? I see similar questions asked on here and obviously I did some research and read my book, but it seems like every explanation contradicts another in some way. There are basically infinite scenarios using these and every example problem/scenario I seem to convince myself it could be both!
Here are some of my understandings of each:
Permutation: Every detail matters and ALL ways of doing something. "Think of permutations as a list."
Combinations: Used for groups. Order and Position DOES NOT matter.

My Confusion: 
a.) If permutations are ALL ways of doing something.. then why does order/position/type matter?
b.) If order does NOT matter with combinations.. why are "Locks" said to have a "combination" when clearly the order does matter with a lock? If the "combination" to unlock something is 1-2-3.. then clearly 1-3-2 would not work. Therefore it seems like order does matter..
c.) If permutations are ALL ways of doing something and if EVERY detail matters.. then why are the number of permutations larger than the number of combinations?
Sorry if I included too much. I'm really struggling with this and every time I think I understand a scenario/problem.. I look at another and have no idea how to do it! I'd greatly appreciate any help. Thank you!
 A: a) this one depends on the question at hand.
Say you have 3 people (A, B, C) and you want to put them in a row. Finding all the arrangements would include
A B C
A C B
B C A
B A C
C A B
C B A
3P3 = 3! = 6 arrangements
but the same three people were in there so if order didn't matter, this would only count as 1 arrangement since in each case there was 1A, 1B, 1C.
that's how a combination is different
$3\choose3$ = 1 arrangement
b) You're right that's clever. Locks don't have permutations or combinations actually
you can't say a lock has 10P3 = 10 x 9 x 8
A lock has different numbers for each position
so n(S) = 10x10x10
c) The combination is just the number of choices provided we are not ordering the items but choosing a certain amount of them and grouping them together. A permutation is all the ways of arranging all the combinations into specific orders like in a)
A question on combinations.
If there are 30 people in a class and you need to pick 2 people to clean up at the end of the day. How many ways can you choose those people. This is one where order doesn't matter because picking Eddie and Fred is the same as Fred and Eddie
n(S) = $30\choose2$ = 435 (not 30P2 = 870. Note that 870 is double 435 so see how in this case you would have had 870 full arrangements but only half of them are valid choices. Why is that? Obviously because for every Fred and Eddie there was a Fred Eddie)
If you are choosing 3 people it gets worse because
30P3 = 24 360 but $30\choose3$ is 4 060. Clearly you cannot just divide the permutation by 3 to get 4060. You need to divide it by 3! because there will be 
Fred, Eddie, Freddie. 
Fred, Freddie, Eddie.
Eddie, Fred, Freddie.
Eddie, Freddie, Fred.
Freddie, Fred, Eddie.
Freddie, Eddie, Fred.
$A\chooseB$ = $A\PermuteB$ / B!
A: Thank you for question and explanations above.  Below is the way I try to remember it.
'Does not matter' actually means 'no replacement' AND 'no repetition within arrangement'
[Note: In other words $[1,2]$ or $[2,1]$ does not matter, we can pick only one of those two arrangements.]
Briefly:


*

*Lock: All combinations allowed (with replacement AND repetition both)

*Permutation: NO replacement AND repetition (or rearrangement of the same set) allowed

*Combination: NO replacement AND NO repetition (or rearrangement of the same set)


Example:
Code to a lock with 2-digits: --  --  using only three digits: 7,8 and 9
[Note: Here we have a total of 3 digits to work with: 7,8 and 9]
1. Lock:
Question: 
Find all possible 2-digit codes for the lock
Answer: 
$3 \times 3 = 9$ 
Possible codes for the lock:


*

*1,1 

*1,2

*1,3

*2,1

*2,2

*2,3

*3,1

*3,2

*3,3


Explanation:
Because here we assume that our lock could have "replacement and repetition". Thus the possibilities of arranging the code could be anything.
2. Permutation:
Now let's assume that we have the same lock with 2-digit code, however now if a digit is used once in the 2-digit-code, then it cannot be used again i.e. NO replacement.
Question: 
Find 2-digit code with NO replacement AND with repetition.
Answer: 
$3 \times 2 = 6$
or
$^{n}\textrm{P}_k  = \frac{n!}{(n-k)!} =\frac{3!}{(3-2)!}   = 6$
Possible codes for the lock:


*

*1,2

*1,3

*2,1

*2,3

*3,1

*3,2


[Note: Removed 1,1; 2,2; and 3,3 because repetition is NOT allowed i.e. repetition of the same digit within a code]
Explanation:
Because let's say we use the digit '1' in the first place of the 2-digit code, then we have only 2 out of 3 digits remaining for the second place of the 2-digit code. Thus, one less number for each subsequent digit of the code. 
3. Combination
Here, we add additional constraints on top of the Permutation i.e. NO replacement and NO repetition.
Question: 
Find 2-digit code with NO replacement and NO repetition
Answer: 
$^{n}\textrm{C}_r  = \frac{n!}{(r! * (n-r)!} =\frac{3!}{2!*(3-2)!}   = 3$
Possible codes for the lock:


*

*1,2

*1,3

*2,3


[Note-1: Removed 1,1; 2,2 and 3,3 because replacement is NOT allowed.]
[Note-2: Removed 2,1; 3,2 and 3,1 because repetition is NOT allowed. Because 2,1 or 1,2 is does not matter and we can pick only one of those two. The ORDER DOES NOT MATTER!]
Explanation:
Here, we start off with 3 total digits that we can use. However, we have two constraints. The first constraint is that we cannot replace or re-use a digit once it is used (same as permutation above). Additionally, now we cannot repeat an arrangement of digits if they ALL are the same digits.
