In a family, grandfather has 3 sons and 24 grandsons. First son has x children and second son has x + 1 children. If children of different parents shake hands with each other, (i.e. no children of the same parents shake hands among themselves), find the maximum possible number of hand shakes that can take place.

Now 1^st son has x children, 2^nd has x+1 children. So, 3^rd will be having (24-(2x+1))children, i.e. (23-2x) children. Total number of handshakes is ²⁴C₂ = 276 But it includes cases in which children of same parents shake hands among themselves. So, possible number of handshakes = 276 - x (Handshakes among children of same parents) According to question, it should be maximum, i.e. handshakes among children of same parents should be minimum. Handshakes among children of same parents (y) y = ^xC₂ + ^(x+1)C₂ + ^(23-2x)C₂ y= 253+3x²-45x Differentiating it w.r.t. x and putting 0, we get x = 7.5 Since the number of children cannot be in fraction, lets take x=8 x+1=9 and remaining children of 3^rd son come out to be 7. So putting the values and solving, we get  y=21+28+36=85 So, maximum possible handshakes are 276-85 = 191 (Correct Answer)