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Mathematics Question on permutations and combinations

There are 1515 points in a plane, no three of which are in a straight line, except 66, all of which are in a st. line. The number of st. lines which can be drawn by joining them is

A

15C2−6^{15}C_2-6

B

15C2− 6C2^{15}C_2-\,^6C_2

C

15C2− 6C2−1^{15}C_2-\,^6C_2-1

D

15C2− 6C2+1^{15}C_2-\,^6C_2+1

Answer

15C2− 6C2+1^{15}C_2-\,^6C_2+1

Explanation

Solution

Reqd. no. of st. lines =15C2− 6C2+1= ^{15}C_{2}-\,^{6}C_{2}+1 [∵\because join of two pts. from a st. line and 66 pts. which are collinear given only st. line in place of 6c2^{6}c_{2}]