Concept: The question tests on the application of counting
Let the persons selected be X, Y and Z as shown below
|- - - X - - - Y - - - Z - - -|
Let a denote the number of persons sitting to the left of X.
Let b denote the number of persons between X and Y.
Let c denote the number of persons between Y and Z.
Let d denote the number of persons to the right of Z.
Therefore a + b + c + d = n - 3 as X,Y and Z are not included in a,b,c and d.
a and d can be 0 assuming that the 1st person is X and the last person is Z.
b and c cannot be 0 as then X and Y or Y and Z would become consecutive. To avoid this we assume b and c to be another variable + 1
Let us assume that b = p + 1 and c = q + 1, where now p and q can be equal to 0.
Now a + p + 1 + q + 1 + d = n - 3
Therefore a + p + q + d = n - 5
If the number of non-negative integer solutions for the equation x1+x2 + ..+ xn = nx1+x2 + ..+ xn = n, then the number of ways the distribution can be done is
n+r−1Cr−1. In this case, value of any variable can be zero.
Here r = 4 and n = n - 5
Therefore Pn = n−5+4−1C4−1=n−2C3
Now the same principle applies when seated in a circle, but it is a little bit more complex. Let us understand it with numbers.
Suppose n = 5 and the same condition is to be applied.
Total number of ways where no 2 are adjacent = Total ways of choosing 3 people - Total ways where all are adjacent - Total ways where 2 are adjacent and 1 is separate
(a) Total ways of choosing 3 people = 5C3. Therefore for n people = nC3
(b) Total ways where all are adjacent = 5 ways. Therefore for n people it is n ways
(c) Total ways where 2 are adjacent and 1 is separate = Any 2 adjacent people can be selected in 5 ways
Any person non adjacent to these 2 can be chosen in 1 way = 5 - 4 ways (this is constant for any value we take for any number)
So total number of ways = 5 * (5 - 4) ways. For n people it is n * (n - 4) ways
Qn = nC3 - n - n(n - 4)
Given that Pn - Qn = 6
4n - 3n = 10
n = 10