How many necklaces of 12 beads can be made from 18 beads of various colours?

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

When All S are taken together, then ASSASSINATION has 10 letters .
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So, 10 letters in total can be arranged in 10 ! ways .
[? All 'S' are considered as 1]

But, here are 3 'A' and 2'I' and 2 'N' .
? Required number of ways = 10 ! / (3 ! x 2 ! x 2 !) = 151200

The first marble can be put into the pockets in 4 ways, so can the second and third.
Thus, the number of ways in which the child can put the marbles = 4 x 4 x 4 = 64 ways .

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

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

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