In how many ways, a cricket team of 11 players can be made from 15 players, if a particulars player is always chosen?

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

After fixing up one boy on the table, the remaining can be arranged in 4 ! ways, but boys and girls have to be alternate. There will be 5 places, one place each between two boys. These 5 place can be filled by 5 girls in 5 ! ways .
Hence, by the principle of multiplication, the required number of ways = 4 ! x 5 ! = 2880

Total persons on the circular table
= 20 guests + 1 host = 21
They can be seated in (21 - 1) ! = 20 ! ways .

Number of arrangements of beads = (12 - 1) !, but it is not mentioned that either it is clockwise or anti - clockwise .
So, required number of arrangements = [(12 - 1)! ] / 2 = 11 !/2

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

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 straight lines
=ⁿC₂ - ^mC₂ + 1

Here, n = 10, m = 5
= ¹⁰C₂ - ⁵C₂ + 1
= 45 - 10 + 1 = 36

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