The question is in the title. I have no idea how to do this one. I was trying to rewrite $n^2$ as $\binom {n}{1} \times \binom {n}{1}$. I was going to say the story is, suppose there are $n$ boys and $n$ girls. How many ways there are to pick $1$ boy and $1$ girl, which is the LHS. But then I don't have any idea how to prove the RHS.
Give a combinatorial proof for $n^2 = n(n-1)+n$
combinatorial-proofs
Best Answer
If we have two people that need to choose from $n$ items (not necessarily distinct),
then there are $n \cdot n = n^{2}$ possibilities.
If we instead count the distinct possibilities first we subtract one for the second person's choice, which gives $n(n-1)$.
Note that since we have $n$ items, there are exactly $n$ possibilities where they do choose the same item.
These are independent choices with the union being the first counting method:
i.e. Distinct + Not distinct = Not necessarily distinct.
Hence we add them and get $n^{2}=n(n-1)+n$.