Answer
$C(5, 3)=10$
The possible combinations are:
$abc,abd,abe,acd,ace,ade\\
bcd,bce,bde\\cde$
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,
$C(5, 3)=\dfrac{5!}{(5-3)!3!}=\dfrac{5!}{2!3!}=10$
The list of all combinationsL
$abc,abd,abe,acd,ace,ade\\bcd,bce,bde\\cde$
