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

Hint: In the above question, first of all we will assume that after marrying Sohan and Geeta have ‘y’ children together. Then, fights will take place among ‘x’ children of Sohan with (x+1) children of Geeta or we can say ‘y’ children of both. We can form a relation between ‘x’ and ‘y’ using the given condition as $x+y+\left( x+1 \right)=24$. Then we can use the concept of combination to find the total number of fights.

Complete step-by-step answer:
We have been given that Sohan has ‘x’ children and (x+1) children of Geeta with her first wife and first husband respectively. After marrying Sohan with Geeta they have a total of 24 children.
Let us assume that after marrying Sohan with Geeta, they have ‘y’ children.
Then, total number of children $=x+y+\left( x+1 \right)$
According to the question,

& x+x+1+y=24 \\\ & \Rightarrow 2x+y+1=24 \\\ & \Rightarrow 2x+y=23 \\\ & \Rightarrow y=23-2x.....(1) \\\ \end{aligned}$$ Now let the number of fights be ‘z’. A child from ‘x’ will fight with the one with (x+1) or from ‘y’. $$\Rightarrow $$Number of fights $${{=}^{x}}{{C}_{1}}{{\times }^{x+1}}{{C}_{1}}{{+}^{x}}{{C}_{1}}{{\times }^{y}}{{C}_{1}}$$ Since we select one child from x, (x+1) and y, so we take $$^{x}{{C}_{1}}{{,}^{x+1}}{{C}_{1}}{{,}^{y}}{{C}_{1}}$$. Similarly, (x+1) will fight with ‘x’ or ‘y’ and ‘y’ will fight with ‘x’ or (x+1). Total number of fights, $$z{{=}^{x}}{{C}_{1}}{{\times }^{x+1}}{{C}_{1}}{{+}^{x}}{{C}_{1}}{{\times }^{y}}{{C}_{1}}{{+}^{x+1}}{{C}_{1}}{{\times }^{y}}{{C}_{1}}$$ We know that $$^{n}{{C}_{1}}=n$$. $$\begin{aligned} & \Rightarrow z=x\times (x+1)+x\times y+(x+1)\times y \\\ & \Rightarrow z={{x}^{2}}+x+xy+xy+y \\\ & \Rightarrow z={{x}^{2}}+2xy+x+y \\\ \end{aligned}$$ On substituting the value of ‘y’ from equation (1) we get as follows: $$\begin{aligned} & \Rightarrow z={{x}^{2}}+2x(23-2x)+x+23-2x \\\ & \Rightarrow z={{x}^{2}}+46x-4{{x}^{2}}+x+23-2x \\\ \end{aligned}$$ On rearranging the terms, we get as follows: $$\begin{aligned} & \Rightarrow z={{x}^{2}}-4{{x}^{2}}+46x+x-2x-23 \\\ & \Rightarrow z=-3{{x}^{2}}+45x+23 \\\ \end{aligned}$$ Since ‘x’ is a real number, the discriminant value ‘D’ will be greater than or equal to zero for the above quadratic equation. $$\Rightarrow z=-3{{x}^{2}}+45x+(23-z)=0$$ We know that for $$a{{x}^{2}}+bx+c=0$$, $$D={{b}^{2}}-4ac$$. We have $$a=-3,b=45,c=(23-z)$$ $$\begin{aligned} & \Rightarrow D={{\left( 45 \right)}^{2}}-4\left( -3 \right)\left( 23-z \right)\ge 0 \\\ & \Rightarrow 2025+12\left( 23-z \right)\ge 0 \\\ & \Rightarrow 2025+12\times 23-12z\ge 0 \\\ & \Rightarrow 2025+276-12z\ge 0 \\\ & \Rightarrow 2301\ge 12z \\\ \end{aligned}$$ On dividing the inequality by 12, we get as follows: $$\begin{aligned} & \Rightarrow \dfrac{2301}{12}\ge z \\\ & \Rightarrow z\le 191.75 \\\ \end{aligned}$$ Since z must be a whole number, so the minimum value for z is 191. Hence, it is proved that the maximum possible number of fights that can take place is 191. Note: Just be careful while calculating as there is a chance that you might make a sign mistake while finding $$D\ge 0$$. In this question we used the concept that the number of ways of selecting 1 thing from ‘n’ different things is equal to $$^{n}{{C}_{1}}$$ which is equal to ‘n’.