Question

Sohan has $x$ children by his first wife. Geeta has $(x+1)$ children by her first husband. They marry and have children of their own. The whole family has $24$ children. Assuming that two children of the same parents do not fight, prove that the maximum possible number of fights that can take place is $191$.

Answer 1

The whole family has $24$ children. Children of Geeta and Sohan are $24-x-(x+1)=23-2x$ in number. When Sohan's child fights with a Geeta's child, there are $x(x+1)$ fights. When Sohan's child fights with Sohan and Geeta's children, there are $x(23-2x)$ fights. Again, when Geeta's child fights with Sohan and Geeta's children, there are $(x+1)(23-2x)$ fights. Therefore, total number of fights is
$F\left(x\right)=x(x+1)+x(23-2x)+(x+1)(23-2x)$
$=23+45x-3x^2$
$=3[\frac{23}{3}+\frac{225}{4}-{(x-\frac{15}{2})}^2]$
For $F\left(x\right)$ to be maximum,
$x-\frac{15}{2}=0$,
i.e.,
$x=7.5$
Since $x$ is an integer, therefore, $x=7$ or $8$. For both $x=7$ and $x=8$. Total number of fights is $F\left(x\right)=191$.