Answer
The combinations are: $\text{abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde}$.
$C(5, 3)=10$ combinations.
Work Step by Step
There are $10$ combinations, namely:
$\text{abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde}$
We know that $C(n,r)=\dfrac{n!}{(n-r)!r!}$
Hence,
$C(5,3)=\dfrac{5!}{(5-3)!\cdot3!}=\dfrac{5\cdot4\cdot3!}{2\cdot1\cdot3!}=\dfrac{5\cdot4}{2}=10$.
