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I know that a circular permutation will be used here but I don't know how because of the number of available chairs. I just can't seem to get the concept of it. Any help will be appreciated, and citing sources is good, too. Thanks in advance. :)

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No. of ways of choosing $3$ friends out of $5$ is $\displaystyle \binom{5}{3}$.

No. of ways of seating those $3$ around the table is $\displaystyle (3-1)!$

Therefore, total ways are $${\binom{5}{3}\cdot (3-1)!}$$

Check this for brief circular permutaion explanation or this for an ellaborate one.

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    Doesn't he want the concept? – 2017-01-05
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    The concept can be extrapolated from the answer fairly easily. – 2017-01-05
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    I'd say true to that, Ian. – 2017-01-05
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    By the way, thank you, Wicked Poison for the answer and sources. It helped a lot. – 2017-01-05
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I'll try to explain the concept.

If you line $n$ people up, there's a leftmost person and a rightmost person. Every person's position in the line is uniquely determined, and there are hence $n!$ ways to line them up.

Putting these same $n$ people around a circular table is the same as putting them in a line, except that the leftmost person is free to pick her seat anywhere at the table. The others, though, must "line up" around the table as they were, with her at the start. Hence, it's essentially like you're lining up only $n-1$ people, and there are $(n-1)!$ ways to do that.

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    I got your point, brother. Thanks for the feedback! – 2017-01-05
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This is meant to be a more general and more elaborate answer.

If you have $m$ friends and you want to seat them at a circular table with only $n$ ($n \le m$) chairs, you first need to decide which people actually sit at the table. Now you want to know how many ways there are to choose $n$ of your $m$ friends, that's what the binomial coefficient is for. There are $\binom{m}{n}$ ways to choose $n$ objects out of a set $m$ elements where the order does not matter.

Now to figuring out how many ways there are to seat your $n$ friends on as many chairs. Take any of your friends and seat them on any of the chairs, there are $n$ chairs and as many ways of seating your first friend. For the second friend there are now only $n-1$ ways and so on. There are $n!$ possible ways of seating $n$ people on the $n$ chairs.

But the table is circular! That means currently some of your "different" ways of seating people, are actually the same, if you just rotate the table. Namely for every permutation there are $n$ equal permutations (itself included) that can be reached from the original one just by rotating the table a bit (or not at all). So currently you have $n$ times as many "different" ways of seating $n$ people at the table than you need.

From this follows that the actual number of possible ways of seating $n$ people at a circular table with $n$ chairs is $\frac{n!}{n}=(n-1)!$

Since there are $m$ people to choose from, the number of possible ways of seating $m$ people at a circular table with $n$ chairs is: $$ \binom{m}{n} \cdot (n-1)! $$

Or in your case: $$ \binom{5}{3} \cdot (3-1)!=20 $$