There are an algebraic way and a combinatorial method to approach them. I will do the combinatorial method for (2) and the algebraic method for (3):
(2): Let us count the size of the set M={(c, A)| a is in [1..n], c is in A, and A is a subset of [1..n] with size r} in two different ways, where [1..n] is the set of the first n positive integers. It is clear that there are there C(n,r) subsets of size r, that is, there are r choices for A. For each A, there are r choices for c as c must be in A. So the size of M is C(n,r)*r. On the other side, each of the n integers in [1..n] can be c, so there are n choices for c. For each c, the number of subsets containing c is C(n-1,r-1) because any subset of size r-1 not containing c can be extended to a subset of size r by joining c into it. So the size of M is n*C(n-1,r-1). Now we conclude that r*C(n,r)=n*C(n-1,r-1).
(3) C(2n,n)=(2n)!/(n!*n!)=[(2n-1)!(2n)]/[(n-1)!*n*n!] (Noting that (2n)!=(2n-1)!*(2n) and n!=(n-1)!*n)
=2(2n-1)!/[(n-1)!*n!] (Cancelling n in the denominator and numerator)
=2C(2n-1, n-1)
Therefore, 2C(2n-1, n-1) = C(2n,n).
