Proving using a combinatorial argument?

Question

$\binom{m+n}{2} = \binom{n}{2} + \binom{m}{2}+ \binom{n}{1}\binom{m}{1}$

I know that the RHS means the ways you can choose $2$ numbers from $\{2, 3, ... ,n+m\}$. I am a little confused on how the LHS equates that.

Answer 1

Consider a bag with $m$ blue balls and $n$ red balls, each of which uniquely labeled.

Count the number of ways that you can select two balls simultaneously without replacement

• Directly

• Breaking into cases based on whether both balls were blue, both balls were red, or you got one of each

Answer 2

From $m$ maths distinct and $n$ number of distinct science books $2$ books can be selected is $={m+n \choose 2}$ . $$ $$ Another way to select either $2$ maths , $2$ science and $1$ maths and science each is $={m \choose 2}+{n \choose 2} +{m \choose 1}{n \choose 1}$ $$. $$ Hence ${m+n \choose 2}={m \choose 2}+{n \choose 2}+{m \choose 1}{n \choose 1}$