by noah Thu May 19, 2011 12:10 pm
Yup - there's a pretty efficient way to solve this. First, let's interpret the new condition.
If P must be assigned to any subcommittee that F is assigned to, as well as any that G is assigned to, it seems that P must be the one that serves on all 3. If you're not convinced, try putting M on all 3 and then P with F and P with G - who's left? H and I, and they can't be together.
So, we've got P in all 3, with F and G a top two of them:
_ _ _
F G __
P P P
-- -- --
Now, who's left? We have to place H and I in different places, M and one more mystery character. So, this could look like this:
(H I M ?)
_ _ _
F G _
P P P
-- -- --
This should make it pretty easy to evaluate the answer choices:
(A) can't be true because if M is with F and G, then H and I are stuck together.
(B) can't be true - M could be with either H or I, but not with both. There's only one column that still has 2 available slots.
(C) is correct - H could be in the F and G subcommittees, leaving M and I together.
You'd pull the trigger at this point. But, for review's sake:
(D) can't work - we only have one mystery character, this would require two extra Fs (beyond the one that's already assigned)
(E) is similar to (D) - there aren't enough mystery characters for this option.
The key to solving this efficiently is to follow the inference chain and then show yourself "who's left" so that you can quickly evaluate the answer choices.
Does that help?