Finding the Least Possible Value of $a$ for a Perfectly Balanced School Club

[anemone]

In a school we have $a$ girl and $a$ boy students with $a>2013$. We know that the number of ways we can choose a club consisting of 6 girls and 5 boys is a square number. What is the least possible value of $a$?

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[anemone]

Congratulations to the following members for their correct solutions::)

1. kaliprasad
2. lfdahl
3. Opalg
4. MarkFL

Solution from MarkFL:

Spoiler

The number $N$ of ways to choose 6 from $a$ and 5 from $a$ is given by:

\(\displaystyle N={a \choose 6}\cdot{a \choose 5}=\left(\frac{(a-4)(a-3)(a-2)(a-1)a}{120}\right)^2\cdot\frac{a-5}{6}\)

In order for $N$ to be a square, we require \(\displaystyle \frac{a-5}{6}\) to be a square. Given that $2013<a$ and:

\(\displaystyle 18^2<\frac{2013-5}{6}<19^2\)

We need to solve:

\(\displaystyle \frac{a-5}{6}=19^2=361\)

\(\displaystyle a-5=2166\)

\(\displaystyle a=2171\)

To check:

\(\displaystyle N=7600981113251526^2\)