In how many different ways, 5 boys and 5 girls can sit on a circular table, so that the boys and girls are alternate?

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 filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 2⁵ = 32
Two adjacent boxes with blue balls can be obtained in 4 ways, i.e., (12), (23), (34) and (45). Three adjacent boxes with blue balls can be obtained in 3 ways i.e., (123), (234) and (345). Four adjacent boxes with blue balls can be obtained in 2 ways i.e., (1234) and (2345) and five boxes with blue balls can be got in 1 way.

Hence, the total number of ways of filling the boxes such that adjacent boxes have blue balls
= (4 + 3 + 2 + 1)
= 10

Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue balls
= 32 - 10
= 22

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

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