**Concept: The question tests on the application of counting**

**Solution:**

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

n−2C3−[nC3−n−n(n−4)]=6n−2C3−[nC3−n−n(n−4)]=6

(n−2)!/(n−2−3)!∗3!−[n!/(n−3)!∗3!−n−n2+4n]=6

(n−2)!/(n−5)!∗6−[n!/(n−3)!∗6−n2+3n]=6

(n−2)(n−3)(n−4)(n−5)!/(n−5)!∗6−[n(n−1)(n−2)(n−3)!/(n−3)!∗6]+n2−3n=6

(n−2)(n−3)(n−4)/6−[n(n−1)(n−2)/6]+n2−3n=6

(n−2)/6[(n−3)(n−4)−[n(n−1)]+n2−3n=6

(n−2)/6[n2−7n+12–(n2–n)]+n2−3n=6

(n−2)/6[n2−7n+12−n2+n]+n2–3n=6

(n−2)/6(−6n+12)+n2–3n=6

(n−2)(−n+2)+n2–3n=6

–n2+2n+2n−4+n2–3n=6

4n - 3n = 10

n = 10**Option C**