Answer
$C(6, 3)=20$
The posible combinations are:
$123,124,125,126,134,135,136,145,146,156\\
234,235,236,245,246,256\\
345,346,356\\
456$
Work Step by Step
If we want to choose $r$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $C(n, r)=\frac{n!}{(n-r)!r!}$ ways.
Hence,
$C(6, 3)=\dfrac{6!}{(6-3)!3!}=\dfrac{6!}{3!3!}=20$
The list of all possible combinations:
$123,124,125,126,134,135,136,145,146,156\\
234,235,236,245,246,256\\
345,346,356\\
456$
