Let Pn denotes the number of ways of selecting 3 people out of ‘n’ sit

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


\(^{n+r−1}C_{r−1}\). In this case, value of any variable can be zero.


Here r = 4 and n = n - 5

Therefore Pn = \(^{n − 5 + 4−1}C_{4−1} = ^{n−2}C_3\)



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


\(^{n−2}C_3 - [^{n}C_3 - n - n(n - 4)] = 6\)

\(\frac{(n - 2)!}{(n - 2 - 3)! * 3!} - [\frac{n!}{(n - 3)! * 3!} - n - n^2 + 4n] = 6\)

\(\frac{(n - 2)!}{(n - 5)! * 6} - [\frac{n!}{(n - 3)! * 6} - n^2 +3n] = 6\)

\(\frac{(n - 2)(n - 3)(n - 4)(n -5)!}{(n - 5)! * 6} - [\frac{n(n -1)(n -2) (n - 3)!}{(n - 3)! * 6}] + n^2 - 3n = 6\)

\(\frac{(n - 2)(n - 3)(n - 4)}{6} - [\frac{n(n -1)(n -2) }{6}] + n^2 - 3n = 6\)

\(\frac{(n - 2)}{6} [(n -3)(n -4) - [n(n - 1)] + n^2 - 3n = 6\)

\(\frac{(n - 2)}{6} [n^2 - 7n + 12 – (n^2 – n)] + n^2 - 3n = 6\)

\(\frac{(n - 2)}{6} [n^2 - 7n + 12 - n^2 + n] + n^2 – 3n = 6\)

\(\frac{(n - 2)}{6} (- 6n + 12) + n^2 – 3n = 6\)

\( (n - 2)(-n + 2) + n^2 – 3n = 6\)

\( –n^2 + 2n + 2n - 4 + n^2 – 3n = 6\)

4n - 3n = 10

n = 10


Option C

Arun Kumar