This really cannot be a GMAT question. More of a CAT level question. You need to have solved these types of questions or their variants before to be actually be able to solve these types of questions.


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 \(x_1 + x_2 \space + \space ..+ \space x_n \space = \space n\), 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 = \(^{n - 5 + 4−1}C_{4−1} = ^{n - 2}C_{3}\)

and Pn+1 = \(^{n - 5 + 1 + 4−1}C_{4−1} = ^{n - 1}C_{3}\)


Given that Pn+1 - Pn = 15

\(^{n - 1}C_{3} - ^{n - 2}C_{3} = 15\)

\(\frac{(n - 1)!}{(n - 1 - 3)!*3!} - \frac{(n - 2)!}{(n - 2 - 3)!*3!} = 15\)

\(\frac{(n - 1)!}{(n - 4)!*6} - \frac{(n - 2)!}{(n - 5)!*6} = 15\)

\(\frac{(n - 1) * (n - 2) * (n - 3) * (n - 4)!}{(n - 4)!} - \frac{(n - 2) * (n - 3) * (n - 4) * (n - 5!}{(n - 5)!} = 90\)


(n - 2)(n - 3) [n - 1 - n + 4] = 90

\((n^2 - 5n + 6) * 3 = 90\)

\(n^2 - 5n + 6 = 30\)

\(n^2 - 5n - 24 = 0\)

\(n^2 - 8n + 3n - 24 = 0\)

n(n - 8) + 3(n - 8) = 0

(n - 8) (n + 3) = 0


n = 8, -7. Since n cannot be negative, n = 8


Option B

Arun Kumar
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