Two series of a question booklet for an aptitude test are to be given to twelve students. In how many ways can the students be placed in two rows of six each, so that there should be no identical series side by side and that the students sitting one behind the other should have the same series?

The number of ways of selecting 3 men and 3 women out of 6 men and 5 women = ⁶C₃ x ⁵C₃
= 6!/(3! x 3!) + 5/(3! x 2!)
= 20 + 10 = 30

Neglecting the direction of beads in the chain, number of ways of preparing a chain with 5 different coloured beads
= (1/2) x (5 -1)! = (1/2) x 4! = 24/2 = 12

first person will shake hands with 11 other persons. Seconds person will shake hands with 10 other persons, Third person will shake hands with 9 other person and so on.
So , total handshake
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66

OR

2 person are to be chosen from 12 person to have a hand shake
? Possible ways = ¹²C₂ = 12! / 2!(12 - 2)! = 66

Number of ways selecting 1 question from ex-7 = ¹²C₁
Number of ways selecting 1 question from ex-8 = ¹⁸C₁
Number of ways selecting 1 question from ex-9 = ⁹C₁
? Total ways = ¹²C₁ x ¹⁸C₁ x ⁹C₁ = 12 x 18 x 9 = 1944

Required number of member played will be (139 - 1) = 138

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

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

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

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 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 .