Question

permutations : how many ways can 5 people (A-E) sit in a row if A must be to the left of B not not necessarily right next to each other? how many ways if A and B CANNOT sit next to eachother?

Answer 1

how many ways can 5 people (A-E) sit in a row if A must be to the left of B not not necessarily right next to each other?

Maybe list some possibilities, A has to be left of B (the numbers list the number of people we can sit there, starting from 3 since we've removed A and B:

AB3*2*1
A3B*2*1
... this could get tedious fast
3*2*1*AB

Answer 2

A must be left of B - by symmetry these will be half of total : 60 ways

or you can work it like this :-

5! - (4+3+2+1)*3!

Answer 3

Doing it a rather tedious way I get the same result as @ganeshie8:
4! + (6 * 3!)

Answer 4

do you guys have an answer or reasoning for it for how many ways if A and B CANNOT sit next to eachother?

Answer 5

Hmm, that one again looks tedious... there's prob a simpler wat

A3B21
A32B1
A321B
..
B3A21

etc...

Answer 6

Think of A and B as one person, call it Z = AB

So you would have these 4 people

Z, C, D, E

There are 4! = 24 ways to order these 4 people. Double this to get 2*24 = 48 (since AB and BA are different)

So there are 48 ways to order the people if they must be together.

Since they cannot be together, this means that there must be 5! - 48 = 120 - 48 = 72 ways to do it where A and B cannot be together.

Answer 7

oh cmon,! for the first part of the question, A must be always to the left , that makes half of the total arrangements. So simply divide 5! by 2.

Answer 8

wow. that is so true.