Bunuel wrote:
If the President and Vice President must sit next to each other in a row with 4 other members of the Board, how many different seating arrangements are possible?
(A) 120
(B) 240
(C) 300
(D) 360
(E) 720
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:Permutations problems, like this one that appears in the
Veritas Prep Combinatorics and Probability lesson book, require you to determine how many unique items are to be arranged, how many of that number are to be selected, and whether or not any special circumstances are present. In this case, the special circumstance is that two members of the board – the President and Vice President – cannot be separated. This provides a challenge to the math – although there are 6 people to be arranged, there are only 5 unique items, as the President and Vice President form one unit to be moved. For example, while this arrangement is possible:
P VP 1 2 3 4
This one is not:
P 1 2 3 4 VP
Because the President and Vice President move as one unit for their arrangement, they only count as one item to be arranged, and so we truly have 5 items to be arranged.
The formula for arranging N items in a row is that there are N! ways to arrange them, so with 5 items, there are 5! = 120 ways to do so. Or, you can create that formula on your own if memory fails: all 5 “items” are eligible to sit on the left hand side, and for each one that sits at the far left, there are 4 left to sit next to them; for each of those, there are 3 left, and so on, until you have 5 * 4 * 3 * 2 * 1.
There’s one more catch to this problem, however – the President and Vice President can sit on either side of each other: P, VP or VP, P. Because of this, these two arrangements:
P VP 1 2 3 4
VP P 1 2 3 4
Are different, and we need to account for that. Because, in each situation, they could sit P, VP or VP, P, we multiply by 2, and the correct answer is 120*2 = 240, or
answer choice B.If you’ve read this far, you may be interested in a follow-up question to solidify your knowledge of this topic. Say that, instead of the President and Vice President with the four board members, we had the Three Stooges (Larry, Curly, and Moe). How would that change the answer?
In this case, we’d still start with 5! = 120, but for the ways in which they can sit amongst themselves, there are not 3 but actually 3! ways that they can sit, as within their group there are 3 “items” to be arranged in a row. In this case, there would be 120*3!, or 720 possible seating arrangements.
With most difficult Combinatorics problems, your toughest responsibility will be finding the unique situation embedded within the question. Take care to visualize the situation before simply selecting a formula, and you’ll be rewarded for doing so.
Given that I've read everything, I have one question: What would be the most efficient way to calculate how many different seating arrangements are possible if the P and the VP cannot sit together?
I've calculated it like this: 6! - 240 = 480 i.e. All the possible arrangements w/o any restriction minus all the arrangements that are not allowed. Is it correct?
It is not complicated to do, but wanted to check if it is the most efficient way.