The exact meaning of 10!/(3!7!) Assume we toss a coin 10 times, independent of each other. 
Each time we can get Heads (H) or Tails (T) , regardless of whether it is fair or not.
So for example this is one possible outcome: HHHHHHHHHH ie 10 heads. Let's call this a 10-toss sequence. 
The total number of possible 10-toss sequences is 2^10 because each toss has 2 possible outcomes: Heads or Tails.
--The first question is What is the meaning of 10!/(3!7!) ? The answer is that It is the total number of ways, by which we can place the three heads inside the 10-toss sequence. The order by which we place the three heads does not matter. 
--The second question is : What is the total number of possible 3-head sequences?
Is it 10!/(3!7!)? Some say YES. Others, say NO.
A 10-toss sequence with three heads, can be this one: HHH THTHTHT. So three positions are fixed to 'heads'. The remaining seven positions can be H or T. So we have 2^7 possible 7-toss sequences. And the three heads can be anywhere in this 10-toss sequence. 
So an answer can be that the total number of 10-toss sequences with 3 heads is 10!/(3!7!) multiplied by (2^7) . That is in the 10-toss sequence the number of ways by which we can place 3 heads is 10!/(3!7!). Then, we have to say something about the remaining 7-toss sequence. Each position can take Heads or Tails . So 2^7 is the possible number of this 7-toss sequence. 
However, I know that the correct answer is 10!/(3!7!) and that we don't need to multiply by 2^7. But I cannot understand why.  
 A: It seems to me that the most interesting possible meaning to your question is "define a "good" string to be a string that contains the substring $HHH$ at least once.  Thus $HHH$ is good, as is $HHHHHH$ or $THHHTHHHTHHH$.  Compute $G_n$, the number of good strings of length $n$."
To be clear:  I am not at all certain that this is what you mean, but it is an interesting calculation so I will carry it out.
We say a string is "bad" if it isn't good.  Let $B_n$ denote the number of bad strings.  Of course every string is either good or bad so $$B_n+G_n=2^n$$  It is somewhat easier to compute $B_n$. To do it, define some special types of bad strings.  We let $r_n$ denote the number of bad strings of length $n$ which end in $T$. We let $s_n$ denote the number of bad strings of length $n$ which end in $TH$. We let $t_n$ denote the number of bad strings of length $n$ which end in $THH$.  For completeness:  let's say that $H$ is a bad string of type $s$ and that $HH$ is a bad string of type $t$.
Work recursively.  We get a bad sequence of type $r$ by appending a $T$ to a bad sequence of length one less.  Thus $$r_n=B_{n-1}$$  We get a bad sequence of type $s$ by appending an $H$ to a bad sequence of type $r$ of length one less.  Thus $$s_n=r_{n-1}=B_{n-2}$$   We get a bad sequence of type $t$ by appending an $H$ to a bad sequence of type $s$ of  length one less.  Thus $$t_n=s_{n-1}=B_{n-3}$$ 
We deduce that $$B_n=B_{n-1}+B_{n-2}+B_{n-3}$$  so $B$ satisfies the so-called "Tribonacci Recursion".  All that remains is to compute $B_i$ for small $i$.  But $B_1=2,\;B_2=4,\;B_3=7$  Thus $$\{B_n\}=\{2,4,7,13,24,44,81,149,274,504,\cdots\}$$ 
Sanity check:  The good strings of length $4$ are $HHHT,THHH,HHHH$ so $B_4=16-3=13$ as desired.  Similarly, the good strings of length $5$ are $HHHxx,THHHx,TTHHH,HTHHH$ hence $8$ of them.  Thus $B_5=32-8=24$ as desired.
In particular $$G_{10}=2^{10}-B_{10}=1024-504=520$$
A: The reason is that you are over-counting in your interpretation.
Suppose you have $HHHTHTHTHT$. This counts as a three head sequence where the first three are heads. But so do $HHHHHHHHHH$, $HHHTHHHTHT$ and so on.
But these are counted in other places as well, for example, $HHHTHHHTHT$ is counted again when you consider a three head sequence of the fifth, sixth and seventh toss, or a sequence of heads in the first, second, and sixth toss, and so on.
So, to get the correct answer, you just need to find the number of ways you can choose three coins to be heads, i.e., ${{10}\choose{3}}=\frac{10!}{3!7!}$
