Probability flipping a coin 
Can you please explain. How to answer these type of questions with permutation/combination? Is there a way I should approach probability questions? 
 A: The number of ways to choose $2$ things out of $4$ is $\dbinom 4 2 = 6$.  Here are the six ways:
\begin{align}
HHtt \\[15pt]
HtHt \\[15pt]
HttH \\[15pt]
tHHt \\[15pt]
tHtH \\[15pt]
ttHH
\end{align}
All are equally probable.  So figure out what the probability of one of them is and continue from there.
A: How many ways are there to have two heads in four elements?  The answer is "4 choose 2".  What is the probability of getting each of these results?  Let's take a look at one possibility:  HHTT.  The probability of getting exactly this outcome is $\frac12 \frac12 \frac12 \frac12=\frac12^4$.  It becomes clear that the probability for all possibilities with two heads is $(1/2)^4$.  So the answer is $(4 \text{ choose } 2)\cdot\frac12^4.$
I'm not exactly sure why my answer was voted down.  Perhaps the person who did so would like to comment on why?
A: I propose two methods.
The first will follow what you have asked for.
1) By Combinations and Permutations (The multiplication principle)
First understand that by using combinations and permutations you are counting. Determining probabilities through counting means you have to eventually use the fundamental formula for probability. That is,
$$\mathcal{P}(E)=\frac{\mathrm{Number\ of\ Favourable\ Outcomes}}{\mathrm{Number\ of\ Total\ Outcomes}}.$$
These outcomes are all inside a sample space. We must therefore determine this sample space.
The denominator is easy to determine. So lets count. Given two letters "T" and "H", how many combinations of 4 can I make? Well by the multiplication principle, the answer is $2\times 2\times\ 2 \times 2 = 16$. We apply the same motive for the numerator. However, we require to choose 2 things of 4. That is the number of favourable outcomes is $\dbinom 4 2 = 6$ Thus our final probability is given by
$$\mathcal{P}(\mathrm{2\ Heads})=\frac{6}{16}.$$
$$=\frac{3}{8}.$$
2) By Mutually Inclusive/Exclusive Events
First realise that the final outcome generated (E.g. HHHH) is a multi-stage event. This is different from tossing a coin once. Since we are dealing with multi stage events we can use the following fact:
"If the outcomes are mutually inclusive, MULTIPLY their probabilities."
"If the outcomes are mutually exclusive, ADD their probabilities"
For this question, realise that the following outcome required arises through the following:
1) HHTT
2) HTHT
3) TTHH
4) THTH
5) HTTH
6) THHT
EACH final outcome above is mutually inclusive. Thus we multiply their individual multistage probabilities:
$$\mathcal{P}(HHTT) = \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$$
$$=\left(\frac{1}{2}\right)^4$$
$$=\frac{1}{16}.$$
In fact, the probabilities of the outcomes I listed above are all going to be the same! That is,
$$\mathcal{P}(\mathrm{HHTT})=\frac{1}{16}$$
$$\mathcal{P}(\mathrm{HTHT})=\frac{1}{16}$$
$$\mathcal{P}(\mathrm{TTHH})=\frac{1}{16}$$
$$\mathcal{P}(\mathrm{THTH})=\frac{1}{16}$$
$$\mathcal{P}(\mathrm{HTTH})=\frac{1}{16}$$
$$\mathcal{P}(\mathrm{THHT})=\frac{1}{16}$$
Now, all the final outcomes above, collectively are mutually exclusive. Therefore, we add the probabilities. Hence
$$\mathcal{P}(\mathrm{2\ Heads})= \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} +\frac{1}{16} + \frac{1}{16}$$
$$=\frac{3}{8}.$$
