Question 1003467
There are {{{6!=1*2*3*4*5*6=720}}} ways that 6 people can be lined up to get on the bus.
(That is because there are 6 ways to choose who is first,
for each of those choices, there are 5 ways to choose who is second,
and so on).


There are {{{4!*2*5=(1*2*3*4)*5=5!=120}}} ways to line up 6 people placing together the 2 people who refuse to be next to each other.
{That is because there are {{{4!}}} ways to arrange the other 4 people,
{{{2}}} ways to arrange the 2 people who refuse to be next to each other,
and {{{5}}} places to insert the problem pair in the line formed by tho other 4).


Since {{{120}}} of the {{{720}}} ways that 6 people can be lined up to get on the bus place together the 2 people refuse to be next to each other,
there are {{{720-120=600}}} ways to line up the 6 people keeping apart the 2 people who refuse to be next to each other.