pemdas ---
First of all, thank you. I am flattered to be cited in a problem.
I have a completely different solution ---- different from anyone else on this page, and different from any of the answer choices.
Take a look at this.
First of all, a point of formal logic. If we have the statement
(not A and not B), the negative or complement of that statement is (
A or B). Similarly, the complement of (
A and B) would be (
not A or not B), and the complement of (
A or B) would be (
not A and not B). The complement of an "and" statement is an "or" statement, and vice versa.
Here, we want to impose the condition
(M & S not adjacent) and (A & R not adjacent).
It's easier to calculate the cases by subtraction, so we have to subtract the complement, which is:
not [(M & S not adjacent) and (A & R not adjacent)] = (M & S adjacent) or (A & R adjacent)
Of course, that "or" region is going to have an overlap that will have to be considered.
So, total number of permutations of six element is, of course, 6! =
720
Consider cases with M & S adjacent. There are ten configurations for M & S
M S _ _ _ _
S M _ _ _ _
_ M S _ _ _
_ S M _ _ _
_ _ M S _ _
_ _ S M _ _
_ _ _ M S _
_ _ _ S M _
_ _ _ _ M S
_ _ _ _ S M
The other four folks can be anywhere (4! = 24), so that's 24*10 =
240 with M & S adjacent, irrespective of A & R.
By symmetry, there would also be
240 with A & R adjacent, irrespective of M & S.
Now, calculate the overlap region, the cases where both M & S are adjacent
and A & R are adjacent.
If M & S are in spaces #1 and #2
M S _ _ _ _ _ (M & S can be in either order ---> x2
then A & R can appear adjacent in #3 & #4, or #4 & #5, or #5 & 6 ----> x3
and A & R can appear in either order ----> x 2
and Rich & Mitch can be in either order in the remaining two spaces ---> x2
2x3x2x2 =
24 possibilities
If M & S are in spaces #2 and #3
_ M S _ _ _ _ (M & S can be in either order ---> x2
then A & R can appear adjacent in #4 & #5, or #5 & 6 ----> x2
and A & R can appear in either order ----> x 2
and Rich & Mitch can be in either order in the remaining two spaces ---> x2
2x2x2x2 =
16 possibilities
If M & S are in spaces #3 and #4
_ _ M S _ _ _ (M & S can be in either order ---> x2
then A & R can appear adjacent in #1 & #2, or #5 & 6 ----> x2
and A & R can appear in either order ----> x 2
and Rich & Mitch can be in either order in the remaining two spaces ---> x2
2x2x2x2 =
16 possibilities
If M & S are in spaces #4 and #5
==> symmetrical to when they're in spaces #2 & #3 ===>
16 possibilities
If M & S are in spaces #5 and #6
==> symmetrical to when they're in spaces #1 & #2 ===>
24 possibilities
Total overlap region = 24 + 16 +16 + 16 + 24 = 96
This overlap has been counted twice, as part of the 240 cases where M&S are adjacent, and also as part of the 240 cases where A&R are adjacent, so it needs to be subtracted.
Total cases where (M & S adjacent)
or (A & R adjacent) = 240 + 240 - 96 = 384
Total cases where (M & S not adjacent)
and (A & R not adjacent) = 720 - 384 =
336
That's my answer, and it's not one of the answer choices.
I believe I have correctly answered the question posed. Is the way I am interpreting the question not as
pemdas intended it to be interpreted?
Are there any questions at all on what I've done?
Mike
) 
