Answer
(i) The combinations of four objects $a, b, c, d$ taken 2 at a time without repetition (note that order does not matter):
a, b
a, c
a, d
b, c
b, d
c, d
(ii) $_4C_2=6$
Work Step by Step
List all the combinations of four objects taken 2 at a time without repetition (note that order does not matter) to have:
a, b
a, c
a, d
b, c
b, d
c, d
There are 6 different combinations possible.
Evaluate $_4C_2$ to have:
$\require{cancel}_4C_2=\dfrac{4!}{2!(4-2)!}
\\_4C_2=\dfrac{4!}{2!(2!)}
\\_4C_2=\dfrac{4(3)\cancel{(2!)}}{2\cancel{(2!)}}
\\_4C_2=\dfrac{12}{2}
\\_4C_2=6$
