Solve for some terms in binomial coefficients

Use the definition of the binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ to establish the following:

Proof. By definition we have,

$\binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{n!}{(n-k)!k!} = \frac{n!}{(n-k)!(n-(n-k))!} = \binom{n}{n-k}. \qquad \blacksquare$
Using part (a) we know $\binom{n}{k} = \binom{n}{n-k}$ . So, if $k = 10$ and $n-k = 7$ , then $n = 17$ .
Again, by part (a) we have $\binom{14}{k} = \binom{14}{k-4}$ implies $k = 14 - (k-4) \implies 2k = 18 \implies k = 9$ .
No. Since $\binom{12}{k} = \binom{12}{k-3}$ implies $k = 12 - (k-3) \implies 2k = 15 \implies k = \frac{15}{2}$ is not an integer.