Odd number of students in odd number of classes In a school there are an odd number of classes, and each class has an odd number of students. We want to choose a school council consisting of one student from each class. Prove that the following are equivalent:
a) There are more ways to form a school council which includes an odd number of boys than ways to form a school council which includes an odd number of girls.
b) There are an odd number of classes which contain more boys than girls.
(British Math Olympiad 2014/15)
I let $b_1>g_1, b_2>g_2, \ldots, b_{2k+1}>g_{2k+1}$ and $b_1'<g_1',b_2'<g_2',\ldots,b_{2l}'<g_{2l}'$, where $k\geq l$. How can we find the number of ways to form a school council which includes an odd number of boys? It is a sum of a lot of terms.
 A: Let there be $k$ classes and let $b_i$ be the number of boys in class $i$. Simularly define $g_i$ for girls. Let $B(k)$ be the number of way we can choose a counsil that consists of odd number of boys and let $G(k)$ be the number of way we can choose a counsil that consists of even number of boys. Using induction we'll prove that:
$$B(k) - G(k) = (b_1-g_1)(b_2-g_2)(b_3-g_3)...(b_k-g_k) \text{ if k is odd}$$
$$G(k) - B(k) = (b_1-g_1)(b_2-g_2)(b_3-g_3)...(b_k-g_k) \text{ if k is even}$$
Induction Base
Let $k=1$, then it's obvious that $B(1) = b_1$ and $G(1) = g_1$, hence $B(1) - G(1) = b_1 - g_1$
If $k=2$ then $B(2) = b_1\cdot g_2 + g_1 \cdot b_2$ and $G(2) = b_1 \cdot b_2 + g_1 \cdot g_2$. Then obviously $B(2) - G(2) = (b_1 - g_1)(b_2 - g_2)$ 
Inductive Hypothesis
Let for $\forall s<k$ hold 
$$B(s) - G(s) = (b_1-g_1)(b_2-g_2)(b_3-g_3)...(b_k-g_k) \text{ if s is odd}$$
$$G(s) - B(s) = (b_1-g_1)(b_2-g_2)(b_3-g_3)...(b_k-g_k) \text{ if s is even}$$
Inductive Step
There are two ways to get a counsil that has $k$ members and odd number of boys. First it to add a new boy to a $k-1$ member counsil that has even number of boys and the other is to add a girl to a $k-1$ member counsil that has odd number of boys. In other words
$$B(k) = b_k \cdot G(k-1) + g_k \cdot B(k-1)$$
Simularly to find the number of $k$ member counsils that have even number of boys, we must add boy to a $k-1$ member counsil with odd number boys or a girl to a $k-1$ member counsil with even number of boys. In other words:
$$G(k) = b_k\cdot B(k-1) + g_k \cdot G(k-1)$$
Now we have:
$$B(k) - G(k) = b_k(G(k-1) - B(k-1)) + g_k(B(k_1) - G(k-1)) = (g_k - b_k)(B(k-1) - G(k-1))$$
So if $k$ is even:
$$G(k) - B(k) = (b_k - g_k)(B(k-1) - G(k-1)) = (b_k - g_k)(b_{k-1} - g_{k-1})...(b_1 - g_1)$$
If $k$ is odd:
$$B(k) - G(k) = (b_k - g_k)(G(k-1)-B(k-1) = (b_k - g_k)(b_{k-1} - g_{k-1})...(b_1 - g_1)$$
Now since in our case $k$ is odd, $G(k)$ also gives us the number of $k$ member counsils with odd number of girls. 
Now let's assume $a)$ holds, then since $B(k) > G(k)$ odd number of the factors on the RHS are positive, hence in odd number of classes we have more boys than girls. Hence $b)$ holds
Now assum that $b)$ holds, then odd number of the factors are positive, while even number of factors are negative, which means $B(k) - G(k) > 0$, hence $a)$ holds.
From this: $a) \iff b) \text{ Q.E.D}$
A: I'd start with this: A class has $g$ girls and $b$ boys. Obviously, if $b>g$ there are more ways to select a boy to represent the class than there are ways to choose a girl. This means more possible councils exist with a boy from the class than a girl.
Now consider $n$ classes. For the sake of the example say $n=3$. Class $1$ has $b_1$ boys and $g_1$ girls. Class $2$ has $b_2$ boys, and so on.
Let $b_1>g_1$, $g_2>b_2$, and $b_3>g_3$. 
Based on the principle stated above, the greatest number of councils include a boy from Class $1$, a girl from Class $2$ and a boy from Class $3$. This can be easily seen by noting this setup produces $b_1*g_2*b_3$ councils. Substituting any value for its counterpart must result in smaller number (just look at the inequalities).
Based on the logic above the following statement holds: The greatest number of councils are produced when each class is represented by the majority gender.
To explicitly solve your problem: Since every class has an odd number of students, every class has either boys or girls in the majority. Since the are an odd number of classes the council always contains an odd number of boys or an odd number of girls but not both.
The greatest number of councils result from picking from the majority gender in each class. Such councils also have exactly as many boys and girls as the are classes with majority boys and girls, respectively.
Thus there will be more ways to make councils with an odd number of boys iff there are an odd number of classes with majority boys.
