There are 10 points in a plan, out of which 5 are collinear. Find the number of straight lines formed by joining them.

Hexagon has 6 sides .
? n = 6
? Required number of diagonals = ¹⁰C₂ - n
= ⁶C₂ - 6
= 6! / [2!(6 - 2)!] - 6 = 6! / (2!4!) - 6
= (6 x 5 x 4!) / (2 x 4!) - 6
= 15 - 6
= 9

Required number of triangles = ¹²C₃ = (12 x 11 x 10) / 6 = 220

Since, particular player is always chosen. It means that 11 - 1 = 10 players are selected out of the remaining 15 - 1 = 14 players.

? Required number of ways = ¹⁴C₁₀
= 14 ! / (10! x 4!)
= (14 x 13 x 12 x 11) / (4 x 3 x 2 x 1)
= 7 x 13 x 11
= 91 x 11
= 1001

Total number of ways = ⁴C₃ x ⁴C₂ = [4! / {3! x 1!}] x [4! /{2! x 2!}]
= (4 x 4 x 3 x 2 x 1) / (2 x 2)
= 4 x 6
= 24

First, we can select 12 beads out of 18 beads in ¹⁸C₁₂ ways.
Now, these 12 beads can make a necklace in 11! / 2 ways as clockwise and anti-clockwise arrangements are same.

So, required number of ways = [ ¹⁸C₁₂ .11! ] / 2!
= [¹⁸C₆.11! ] / 2!
= [18 x 17 x 16 x 15 x 14 x 13 x 11! ] / [6 x 5 x 4 x 3 x 2 x 1 x 2!]
= [17 x 7 x 13! ] / 2!
= 119 x 13 !/ 2

Here, order is important, then the number of ways in which 12 different balls can be divided between two boys who receives 5 and 7 balls respectively, is = [12! / 5! 7! ] x 2! = 1584

The required number of combinations, when one fruit of each kind is taken
= ⁵C₁ x ⁴C₁ x ³C₁ = 5 x 4 x 3 = 60

Required number of ways
= 2ⁿ - 1 = 2¹⁰ - 1 = 1024 - 1 = 1023

Number of ways in which 8 persons can be selected from 15 persons = ¹⁵C₈
Now, 8 persons can be seated around a circular table in 7! ways.
Now, remaining 7 persons can be seated around a circular table in 6! ways.
? Required number of ways = ¹⁵C₈ x 7! x 6!

When O and A occupy end places. Then, the three letters (M, E, G) can be arranged themselves by 3! = 6 ways and two letters (O, A) can be arranged among themselves in 2! = 2 ways.
? Total number of ways = 6 x 2 = 12