Why do we use nCk when determining numbers of favorable outcomes of coin tosses? After delving back into probability a bit, I'm absolutely stumped as to why we would use nCk to answer the question "What is the probability of getting 3 heads when tossing a fair coin 10 times?" I know that the problem needs to be answered as '#favorable outcomes/ total possible outcomes', but from what I understand about combinations from this source,
(http://ocw.mit.edu/courses/mathematics/18-05-introduction-to-probability-and-statistics-spring-2014/readings/MIT18_05S14_Reading1b.pdf)
the combination function would treat outcomes such as 'HHHTTTTTTT' and 'HTHTHTTTTT" as redundant since each outcome is a subset containing the same elements and is, in effect, 1 combination - yet the answer to the problem says there are 120 possible combinations, each having H occur threes times. I though this would be the same as considering {a,b,c} and {c,b,a} as different permutations of the same combination.  Obviously I'm misunderstanding this, and I haven't been able to find an explanation thus far.
So far, I am stuck on finding an explanation for why combinations would be used when H and T can be used more than once - doesn't this imply that they're not "distinct elements"? According to the MIT pdf, combinations deal with distinct elements belonging to a larger set, and the outcome is the total number of subsets containing k distinct elements without regard to order. Yet in this example the order of H's in the string of flips matters, since getting a different order qualifies that set as a distinct favorable outcome.
Thank you for reading and considering.
 A: It's due to a fundamental misunderstanding of probability. For the simple case:

$\text{Probability of an event E} = \dfrac{\text{Number of cases satisfying E}}{\text{Total number of cases}}$ where all the cases are equally likely.

You totally missed the bold part, and hence thought that the order doesn't matter. You are free to choose any way of dividing the possible outcomes into equally likely cases, but the most reasonable way in your case is to consider each sequence of coin flips as one case. The order clearly matters. "HH" and "HT" and "TH" and "TT" are all equally likely, so one cannot group "HT" and "TH" in one case just because there are the same number of heads and tails. So the total number of cases with $3$ heads is exactly the number of ways to choose $3$ positions in the sequence of $10$ flips to be "H" and the rest to be "T". That comes to $\binom{10}{3}$ ways. The total number of cases is $2^{10}$.
A: More properly they should have said that there are $120$ possible arrangements, or outcomes.   The use of the word "combination" at that point is misleading, as combination usually signifies selection without order.   Here, order of occurrence is actually significant, as each coin is a unique individual.
Consider tossing two coins, and let $X_2$ be the count of heads.  Then 
$$\mathsf P(X_2{=}0) = \mathsf P({\rm TT}) = 1/4 \\ \mathsf P(X_2{=}1) = \mathsf P({\rm HT\cup TH}) = 2/4 \\ \mathsf P(X_2{=}2) = \mathsf P({\rm HH}) = 1/4$$
We can see that the number of (equally-probable) permutations that give the same result is important when evaluating probability mass of the random variable.   Probability is only a simple ratio of counts of distinct outcomes when every distinct outcome is equally likely.

$\{\textsf{No-Heads}, \textsf{One-Head}, \textsf{Three-Heads}\}$ are three distinct outcomes, but $\{TT, HT, TH, HH\}$ are four equally likely distinct outcomes.   That is important!

We count these permutations as: "ways to select places for 'heads' in the arrangement".   Hence the use of the "combination" notation in the binomial distribution's probability measure.   (PS: ${^n{\rm C}_k}$ is also called the binomial coefficient, hence why these distributions are so named.)
Then if $X_n$ is the count of heads in $n$ tosses, we have:
$$\mathsf P(X_n {=} k) = \dfrac{^n{\rm C}_k}{2^n}$$
That is all.

We use  ${^{10}}{\rm C}_3$ to count ways to select a combination of $3$ distinct items out of $10$.   These items happen to be the position we place the 'heads', rather than the coins themselves.   That is why we use the "combination" notation to count permutations of the arrangement.  
(Yes, it is confusing.   Take some time to wrap your brain around it.)
Another way to see it, we are counting permutations of an arrangement of $3$ heads and $7$ tails.   There are $10!$ permutations of ten distinct objects, but these are not all distinct. 
Every arrangement is one of a group of $3!$ which are identical except that different heads are in the same position; but heads are not distinct so these $10!/3!$ groups are each of $3!$ indistinguishable arrangements (wrt the heads).   We can likewise group by the tails.   Thus there are $10!/3!7!$ completely distinct ways to arrange the coins.   This is ${^{10}}{\rm C}_3$, aka: the count of distinct ways to select positions for the three heads in the arrangement.
A: Imagine that you have $10$ balls.  Call the balls $B_1, B_2,...B_{10}$.  The number of sets of $3$ balls which can be made is $10C3$.  A set of three balls could be $\{B_2,B_7,B_9\}$.  Since this is a set, it is equal to $\{B_7,B_2,B_9\}$.  The number of ordered lists you can make without repeating an element is $10P3$.  An example of an ordered list is $(B_2,B_7,B_9)$.  This is distinct from $(B_7,B_2,B_9)$.  The "order does not matter" for the combinations, and "the order does matter" for the permutations, but these terms are often confusing, and I think it is better to think in terms of subsets vs. lists.
In your case, you have 10 coin flips.  These occur in an order, but that fact is a red herring.  We have to change our perspective on the problem a bit. If we want three heads, we can achieve this by picking out a subset of the 10 flips to be heads.  Call the flips $F_1,F_2,...,F_{10}$.  The subset $\{F_2,F_7,F_9\}$ and the subset $\{F_7,F_2,F_9\}$ both refer to the same sequence $THTTTTHTHT$.  This shows that the number of sequences containing exactly three heads is the same as the number of subsets of size three coming from a set with 10 elements.  So the answer is $10\choose3$.
A: Here is a different way of deriving the binomial coefficient formula (from Wilf's generatingfunctionology):
For each nonnegative integer $n$ and integer $k$ with $0\leqslant k\leqslant n$, suppose $f(n,k)$ is the number of $k$-element subsets of $\{1,2,\ldots, n\}$. Since each such subset of $\{1,2,\ldots, n\}$ either contains $n$ or does not, we have the recurrence
$$f(n,k) = f(n-1,k) + f(n-1,k-1) $$
for $n\geqslant 1$, with $f(n,0)=1$ for $n\geqslant 0$. For each $n$, define the generating function
$$B_n(x) = \sum_{k=0}^n f(n,k)x^k. $$
Multiplying each side of the recurrence by $x^k$ and summing over $k\geqslant 1$ we get
$$\sum_{k=1}^n f(n,k)x^k = \sum_{k=1}^n f(n-1,k)x^k + \sum_{k=1}^n f(n-1, k-1)x^k, $$
and hence
$$B_n(x) = 1 = B_{n-1}(x) - 1 + xB_{n-1}(x), $$
so that $$B_n(x) = (1+x)B_{n-1}(x) $$
for $n\geqslant 1$. Since $B_n(x)=f(0,0)=1$, it follows that
$$B_n(x) = (1+x)^n, \; n\geqslant0. $$
Since $B_n$ is a polynomial, it is equal to its Taylor expansion:
$$B_n(x) = \sum_{k=0}^n \frac{B^{(k)}_n(0)}{k!}x^k. $$
It is clear that 
$$B^{(k)}(x) = (n)_k (1+x)^{n-k}, $$
where
$$(n)_k = \prod_{j=0}^{k-1}(n-j) = \frac{n!}{(n-k)!}. $$
Therefore $B^{(k)}(0)=(n)_k$, so that the coefficient of $x^k$ in the Taylor expansion is
$$\frac{(n)_k}{k!} = \frac{n!}{k!(n-k)!}. $$
This is the familiar expression for $\binom nk$, and so 
$$(1+x)^n = \sum_{k=0}^n \binom nk x^k. $$
