In how many ways, can the letters of the word 'ASSASSINATION' be arranged, so that all the S are together?

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

The number of arrangement in which A and B are not together
= Total number of arrangement - Number of arrangement in which A and B are together
= 4! - 3! x 2! = 24 - 12 = 12

3 girls can be be selected out of 5 girls in ⁵C₃ ways. Since, number of boys to be invited is not given, hence out of 4 boys, he can invite them in (2)⁴ ways.
? Required number of ways ⁵C₃ x (2)⁴ = 10 x 16 = 160

Total ways = ⁸C₃ x ⁸C₃
= (8 x 7 x 6)/(3 x 2) x (8 x 7 x 6)/(3 x 2)
= 56 x 56 = 3136

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

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

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

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