Confusion about Combinations and Permutations Order importance is the difference between permutation and combination. For a permutation the  order is important and for a combination isn't.
For example: In permutation, $1,2$ and $3$ have $3 \times 2 \times 1=6$ orders , whereas there is only $1$ order in combination since the order is irrelevant.
Here is the question
Theoretically ,  HHHT, HHTH, HTHH, THHH  should be just the same in combination (order doesn't matter ).
So , Why $4C3 = 4$? ( The combinations with four heads that you can form by tossing a coin four times) Shouldn't it be equal to $1$ because the $4$ orders are just the same ?
Another question , pick $4$ men out of ten men $= 10C4$ which is the same as counting how many combinations could be derived from picking $4$ heads from $10$ tosses of coin. 
1st: MMMM  from  MMMMM   MMMMM  
2nd:HHHHTTTTTT/ HHTTHHTTTT/..... etc 
How could these two questions be the same ? Those units and the way how they are being displayed aren't the same at all . Can anyone of you understand what I am talking about or I am just making things complicated?
The second example is not even picking things out of a sample! 
 A: Okay flipping a coin is a bit of a tricky thing to think about but consider this. Imagine I'm sitting over here and you are where ever you are and I flip a coin 4 times, but I tell you I got 3 heads from my 4 flips. Now I don't tell you the order, because I'm a tad sneaky like that and I ask you what are all the possible ways I could have flipped that coin to get 3 heads? 
Well you think to yourself if I've got 4 places that I have to guess and I know I have to fit 3 heads call them $H_1, H_2, H_3$ and one tails $T$ so you think to yourself well how can I arrange these four options? You decide to list them all out;
$$ (H_1, H_2, H_3, T), (H_1, H_2, T, H_3), (H_1, T, H_2, H_3), (H_1, H_2, T, H_3) $$
$$ (H_1, H_3, H_2, T), (H_1, H_3, T, H_2), (H_1, T, H_3, H_2), (H_1, H_3, T, H_2) $$
$$ (H_2, H_1, H_3, T), (H_2, H_1, T, H_3), (H_2, T, H_1, H_3), (H_2, H_1, T, H_3) $$
$$ (H_2, H_3, H_1, T), (H_2, H_3, T, H_1), (H_2, T, H_3, H_1), (H_2, H_3, T, H_1) $$
$$ (H_3, H_2, H_1, T), (H_3, H_2, T, H_1), (H_3, T, H_2, H_1), (H_3, H_2, T, H_1) $$
$$ (H_3, H_1, H_2, T), (H_3, H_1, T, H_2), (H_3, T, H_1, H_2), (H_3, H_1, T, H_2) $$
(Yep just listed all 24 or you could have just done $^4P_3 =24$) But then you say "hey wait a second? Everything in the first column there is the same!" i.e $ (H_1, H_2, H_3, T)=(H_1, H_3, H_2, T)=(H_2, H_1, H_3, T)=(H_2, H_3, H_1, T)$  and this can be done for every column in the table above so you come to the conclusion that there are 4 different ways I could have flipped three head in 4 coin flips? Namely 
$$ (H,H,H,T), (H,H,T,H), (H,T,H,H), (T,H,H,H) $$
A: Firstly, you got it wrong.
It is with permutations that order matters, not with combinations.
Don't translate each problem to flipping coins. This will confuse you.
$4C3$ means that we pick $3$ elements out of $4$ different elements, where order does not matter.
So, suppose we have $4$ elements $A,B,C,D$
and I have to take $3$ elements. This can in $4$ ways:
$ABC,
ABD,
BCD,
ACD$
