Question

Mr. A has x children by his first wife and Ms. B has x + 1 children by her first husband. They marry and have children of their own. The whole family has 10 children. Assuming that two children of the same parents do not fight, the maximum number of fights that can take place among the children is –

• A. 33
• B. 35
• C. 38
• D. None of these

Answer 1

Answer: (A)

33

Answer 2

No. of children of Mr. A and Ms. B

= 10 – x – (x + 1) = 9 – 2x

Let k = no. of fights between children of different parents

\ k = ¹⁰C₂ – (^xC₂ + ^x+1C₂ + ^9–2xC₂)

= 45 – $\frac{x(x - 1)}{2}$– $\frac{(x + 1)x}{2}$–$\frac{(9 - 2x)(8 - 2x)}{2}$

= $\frac{1}{2}$ [90 – x² + x – x² – x – 72 + 34x – 4x²]

= – 3x² + 17x + 9 = –3 $\left\lbrack x^{2} - \frac{17}{3}x \right\rbrack$+ 9

= – 3 $\left\lbrack x^{2} - \frac{17}{3}x + \frac{289}{36} \right\rbrack$ + 9 + 3 $\left( \frac{289}{36} \right)$

= $\frac{397}{12}$ – 3 $\left( x - \frac{17}{6} \right)^{2}$

\ k is maximum when x = 17/6.

We take x =3.

\Max. Value of k = $\frac{397}{12}$– 3 $\left( 3 - \frac{17}{6} \right)^{2}$= 33.