(a) is just evaluating each:

$\displaystyle C(7,5) = \frac {7!}{5! ( 7 - 5)!} = \frac {7!}{5!2!} = \frac {7 \cdot 6}{2} = 21$

$\displaystyle C(4,4) + C(5,4) + C(6,4) = \frac {4!}{4! (4 - 4)!} + \frac {5!}{4! (5 - 4)!} + \frac {6!}{4! (6 - 4)!}$

$\displaystyle = \frac {4!}{4!} + \frac {5!}{4!} + \frac {6!}{4! 2!}$

$\displaystyle = 1 + 5 + \frac {6 \cdot 5}{2}$

$\displaystyle = 1 + 5 + 15$

$\displaystyle = 21$

(b)

The second method is doing what i did for question two in your previous post, but at the end change C(5,1) to C(5,4) and C(6,2) to C(6,4), why can we do that?

C(n,r) = C(n, n - r)

so, for example, C(6,2) = C(6, 6 - 2) = C(6,4)