Answer
(i) List the combinations of five objects taken 2 at a time without repetition (note that order does not matter) to have:
a, b
a, c
a, d
a, e
b, c
b, d
b, e
c, d
c, e
d, e
(ii) $_5C_2=10$
Work Step by Step
List all the combinations of five objects taken 2 at a time without repetition (note that order does not matter) to have:
a, b
a, c
a, d
a, e
b, c
b, d
b, e
c, d
c, e
d, e
There are 10 different combinations possible.
Evaluate $_5C_2$ to have:
$\require{cancel}_5C_2=\dfrac{5!}{2!(5-2)!}
\\=\dfrac{5!}{2!(3!)}
\\=\dfrac{5\cancel{(4)}2\cancel{(3!)}}{\cancel{2}\cancel{(3!)}}
\\=10$
