Find the values of a and b such that $\binom{2n}{2} = a\binom{n}{2} + b(n^2)$

Question

Find the values of a and b such that $\binom{2n}{2} = a\binom{n}{2} + b(n^2)$

Answer 1

Assuming the question is from a course on combinatorics, we can do the following way:

The LHS is the number of ways we can pick $2$ objects out of $2n$ objects.

Now, let us try to get the RHS: We have two terms, one is a combinatorial term (with just $n$ instead of $2n$ in the term), the other seems to be more an algebraic term. So, let us divide our $2n$ objects into two classes $A$ and $B$. We can pick $2$ objects out of $2n$ objects in two ways: either both objects come from same class, or both objects come from different classes. The first part (which is both come from same class) is $2 \times {n \choose 2}$, and the second part (which is they come from different classes) is $n \times n$. So, the LHS can be equated to $2{n \choose 2} + n^2$. Equating the like terms, we have $a=2, b=1$.

If we go the algebraic way, it is not evident if the solution is unique (unless we solve it completely). Similarly, I am not sure if there are other combinatorial arguments which can lead to different values of $a$ and $b$.

Answer 2

The question is quite simple, convert the binomial coefficients to a more readily solvable form first. $\binom{2n}{2}=\frac{2n(2n-1)}{2}=n(2n-1)$. Similarly, $\binom{n}{2}=\frac{n(n-1)}{2}$. Your equation therefore simplifies to: $n(2n-1)=\frac{an(n-1)}{2}+bn^2$

Equating coefficients and solving simultaneously should allow you to derive the answer.