N Choose K Confusion $$
  \binom{kn}{2} = k\binom{n}{2} + \binom{k}{2} n^2,n,k ≥ 2
$$
Im not sure how to prove algebraically this. Can someone help me?
 A: Since $\binom{m}{2} = \frac{m(m-1)}{2}$, you are seeking to prove:
$$
  \frac{k n (k n -1)}{2} = k \frac{n(n-1)}{2} + n^2 \frac{k(k-1)}{2} \stackrel{\text{expand}}{=} {\color\red{ k \frac{n^2}{2}}}  - k \frac{n}{2} + n^2 \frac{k^2}{2} {\color\red{- n^2 \frac{k}{2}}} \stackrel{\text{factor}}{=} k n \frac{k n -1}{2} 
$$
A: It is just a computation. Recall that $\binom{m}{2}=\frac{m(m-1)}{2}$. Thus the left-hand side is equal to $\frac{kn(kn-1)}{2}$. Compute the right-hand side, and simplify.
A: You are given an urn with $n$ balls of each of $k$ different colors. You ask how many ways are there to choose two balls.
Either you choose two balls of the same color, or you choose two balls of different colors. In the first case, calculate the number of ways to choose two red balls: $\binom{n}{2}$; since there are $k$ different colors, the first quantity is $k\binom{n}{2}$. To calculate the number of combinations of 2 different colored balls, notice that you first choose two different colors (say red and blue), then you count the number of blue balls, then you count the number of red balls. The product is the required quantity, $\binom{k}{2}n^2$.
Adding the two values together gives the number to choose two balls of the same color or two balls of different colors.
