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If the President and Vice President must sit next to each other in a
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31 Aug 2015, 09:21
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Re: If the President and Vice President must sit next to each other in a
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31 Aug 2015, 09:36
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. We need to take President and VP as one unit. Therefore we have 5 people to arrange=5! ways Also, we both President and VP can be arranged in 2 ! ways. Reqd no of ways=5!*2!=240 ways Answer B



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Re: If the President and Vice President must sit next to each other in a
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31 Aug 2015, 14:01
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. Since president and vice president must sit next to each other, they will be considered as single unit. So total number of units to be arranged = 5 Number of ways of arranging president and vice president will be 2. Thus total number of arrangements= 2(5!) = 240 Answer: B



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Re: If the President and Vice President must sit next to each other in a
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01 Sep 2015, 01:57
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. Step1: Make a group of President and Vice President and treat them as one individual Step2: Now we have total 4 Individual and one group of President and vice president and all these 5 can be arranged in a row in (5)! = 120 ways Step3: The group of Vice President and President can exchange the seats among themselves in 2! ways Total Ways of Arrangements = 5!*2! = 120*2 = 240 ways Answer: option B
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If the President and Vice President must sit next to each other in a
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06 Sep 2015, 05:32
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 followup 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.
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Re: If the President and Vice President must sit next to each other in a
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09 Nov 2015, 08:27
Bunuel wrote: 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 followup 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.



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Re: If the President and Vice President must sit next to each other in a
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02 Jan 2017, 21:21
240 is the answer. 4 members + 1 member (which includes president and vice president, since they should always be next to each other ) can be arranged in 5!. Also president and vice president alone can sit in 2! way. Hence , 5! X 2! ways . Sent from my ASUS_Z00LD using GMAT Club Forum mobile app



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Re: If the President and Vice President must sit next to each other in a
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03 Jan 2017, 09:27
anairamitch1804 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?
120
240
300
360
720
Stuck with this problem looking for help. Consider President and Vice President as one person. So overall you have 5 person which can seat in 5! ways. = 120 Now, in each of 5! combinations, president and vice president can interchange themselves in 2 ways. So final answer = 120*2 = 240. Hope it helps.



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Re: If the President and Vice President must sit next to each other in a
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21 Jan 2018, 12:16
Hi All, There are a couple of different ways to approach this question, depending on how you "see" the math involved. Here's a visual way to quickly get to the correct answer... Since the President and VicePresident MUST sit next to one another in a row of 6 people, we could have the following arrangement: P V _ _ _ _ Those remaining 4 spots are essentially a factorial... P V 4 3 2 1 So there are 24 possible arrangements that begin with "P V." Similarly, if we reversed the position of the President and VicePresident, we'd have another 24 arrangements... V P 4 3 2 1 That brings the total to 48 arrangements if the P and the V are in the first two spots. This pattern continues all the way down the line, as long as the P and the V are in two consecutive spots... P V _ _ _ _ _ P V _ _ _ _ _ P V _ _ _ _ _ P V _ _ _ _ _ P V Each option gives us 48 arrangements; since there are 5 options, there are (5)(48) = 240 possible arrangments. Final Answer: GMAT assassins aren't born, they're made, Rich
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