does someone knows how to calculate this?
so far this is what I get,
if all the persons could sit everywhere then 7!=720 options would be posibble. So answer a is eliminated.
If I had for instance A B C D E F, then ABC together could'nt be on the 1st or 2nd or third...up to six options* 2 because could be ( CBA) so that would give me 12 options that ABC or BCA cannot be sit. If I group ABC as a unit then I have 3 out of 4 options that D E F can be sit.
12 * 3C4= 48 so, the option e can be eliminated.
Since all the options that A B C D E F can be sitted is 5!=120 we can eliminate option b. I took this assuming that BC could still sit together. 120 minus the possible options were B could be located under A B C D E F , I get that
AC Cannot sit in certain possitions making a total of 12 not possible options to be sitted for instance, I show below if AC was together six possible options that times 2 because AC could be located as CA.
This gives me 120-12= 108 as the correct answer.
AC B D E F
B AC D E F
D B AC E F
D F B AC G
B D E F AC
AC D E F B
