Question

The number of ways in which ten candidates A1, A2........A10 can be arranged in vertical row(column) when

1)if A1, A2 are next to each other. [9!.2!]

2)if A1 is just above A2. [9!]

3)if A1 is always above A2. [10!/2!]

4)if A1 is always above A2 and A2 is above A3 is.[10!/3!]

5)if A1, A2, A3 sit together in a specified order is. [8!]

Please explain solution in detail for the 3,4,5 specially. And kindly don't close the question for whatsoever reason as i typed it with a lot of hard work.[ Answers in bracket.]

Answer 1

1) if A1 and A2 together consider them as a group. So we have 9 entities which can be arranged 9! ways. Byt A1 and A2 can be arranged among themselves in 2! ways . Hence total number of ways = 9!2!
2)If A1 is just above A2, then A1 can be in positions 1-9 only. For each arrangement position of A2 is fixed and we arrange only the remaining 9 positions. So the number of ways = 9!
5)If A1,A2,A3 sit in a specified order they can be considered as 1 group and no permutations among them, So the number of ways is sinply 8! (1 group + 7 others to be permuted so 8 to be permuted)
3)First we permute all 10 without regard to order of A1 and A2. This is 10! permutations, Out of this we want a specific order for A1 and A2. We actually counted 2! orders for A1 and A2 when we should have had just 1. \Hence we must divide by 2! giving 10!/2!
4) First we permute all 10 without regard to order of A1, A2 and \A3. This is 10! permutations, Out of this we want a specific order for A1,A2 and A3. We actually counted 3! orders for A1, A2 and A3 when we should have had just 1. \Hence we must divide by 3! giving 10!/3!