Confusion in understand combinations flipping a coin

Question

I'm having a hard time visualizing this question:

Assume you flip $3$ coins. What is the probability that you get exactly one head.

$P(\text{head}) = \frac{1}{2}$. $P(\text{Total outcomes}) = 2^3$

I was told in combinations order doesn't matter. The answer is $\frac{{3\choose1}}{2^3}$. Which would be

$\{HTT, TTH, THT\}$. Order matters doesn't it?

Answer 1

Yes, order matters: imagine the analogous problem with just two coin flips.

The three possible outcomes are: two heads, two tails, one of each; if we mistakenly discount the importance of order, we might mistakenly believe each of these three scenarios is equally likely to occur. In other words, that each of the three scenarios occurs with probability $1/3$.

But, try this experiment - flipping a coin twice - many times. See what you observe to be the probability of these scenarios. Over time, it should look like two heads, $1/4$; two tails, $1/4$; one of each, $1/2$. This is because by correctly taking order into account, there are the four scenarios of HH; TT; HT, TH.

A similar line of reasoning applies to the problem about which you ask.

Answer 2

You can see it this way.

HHH- 1

HHT- $\frac{3!}{2!}$ = 3

HTT- $\frac{3!}{2!}$ = 3

TTT- 1

Since the question asks for exaxtly one head, the probability would be $\frac{3}{8}$

You are using combinatorics to find out the possible combinations of HHH, HHT, HTT and TTT. That is why orders matter in this case. Sorry, I don't know how to word it properly.