Quartets and parity There are 229 girls and 271 boys at a school. They are divided into 10 groups of 50 students each, with numbering 1 to 50 in each group. A quartet consists of4 students from 2 different groups so that there are two pairs of students having identical numbers. Show that the number of quartets with an odd number of girls is itself odd.
Any help, please? Would it help to extend to Mod 4? Is the total number of quartets
$${50 \choose 2}\cdot {10 \choose 2} $$???

Let the groups be $G_1,G_2,\ldots,G_{10}$, with respectively $b_1,b_2,\ldots,b_{10}$ boys and $g_1,g_2,\ldots,g_{10}$ girls.
A quartet with an odd number of girls must have either (a) a pair of boys and a boy/girl pair, or (b) a pair of girls and a boy/girl pair.  So, from two groups $G_i$ and $G_j$, the number of quartets with an odd number of girls is precisely $$|b_i-b_j| \times (50-|b_i-b_j|) \equiv \begin{cases} 0 & \text{if } b_i \equiv b_j \pmod 2 \\ 1 & \text{if } b_i \not\equiv b_j \pmod 2 \end{cases} \pmod 2.$$
So it is sufficient to show that there are an odd number of pairs $(i,j)$ for which $b_i$ is odd and $b_j$ is even.
There are an odd number of boys, and an even number of groups, so $\sigma:=\#\{b_i:b_i \text{ is odd}\}$ and $\eta:=\#\{b_i:b_i \text{ is even}\}$ are both odd.  So the number of pairs $(i,j)$ for which $b_i$ is odd and $b_j$ is equal to $\sigma\eta$, which is odd.
