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by 800_or_bust » Wed Jun 15, 2016 11:04 am
Just a question I came up. Figured I'd post it for my fellow would-be GMAT takers to attempt. Solution posted in 2 days.

Seven people - Anna, Bill, Corey, Drew, Eileen, Frank, and George - are being considered for a corporate board of officers, consisting of a president, a treasurer, and a secretary. In how many different ways can this board of officers be arranged if Drew and Eileen cannot serve together in any capacity?
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by diegocml » Thu Jun 16, 2016 6:53 am
I haven't actually seen permutation in my GMAT studies yet. Having said that:

n!/(n-k)!

The fact that Drew and Eileen cannot serve together in any capacity makes me think that n = 6, but if I plug that I get > 6!/(6-3)! = 120, which is not even in the answer choices. Alternatively, if n = 7! I have 210, which I doubt is the correct answer.

I'm clueless, but I think the catch is with Drew and Eileen. Curious to see how to hack this problem.
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by flsh » Thu Jun 16, 2016 7:11 am
800_or_bust wrote:Just a question I came up. Figured I'd post it for my fellow would-be GMAT takers to attempt. Solution posted in 2 days.

Seven people - Anna, Bill, Corey, Drew, Eileen, Frank, and George - are being considered for a corporate board of officers, consisting of a president, a treasurer, and a secretary. In how many different ways can this board of officers be arranged if Drew and Eileen cannot serve together in any capacity?
There is a difference between the members of the board, so we use a permutations:
N = 7P3 - 3P2 * 5P1 = 7!/4! - 3!/1! * 5!/4! = 5*6*7 - 6*5 = 5*6 * (7 - 1) = 30 * 6 = 180.
D is the answer.
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by MartyMurray » Thu Jun 16, 2016 7:21 am
diegocml wrote:I haven't actually seen permutation in my GMAT studies yet. Having said that:

n!/(n-k)!

The fact that Drew and Eileen cannot serve together in any capacity makes me think that n = 6, but if I plug that I get > 6!/(6-3)! = 120, which is not even in the answer choices. Alternatively, if n = 7! I have 210, which I doubt is the correct answer.

I'm clueless, but I think the catch is with Drew and Eileen. Curious to see how to hack this problem.
210 is a good start, and you can tell that it's not the right answer, because 210 is the total possible permutations of all of them, and the answer does not include some of the permutations of all of them. So it has to be less than 210.

You could get the answer this way.

Start with 210, and then subtract all the ones that have Drew and Eileen in them.

The ones that have Drew and Eileen have D, E and 5 other people.

So we fill the slots this way.

D, E, 1 of 5

D, 1 of 5, E

1 of 5, D, E

and so on.

So we have three elements, D, E, 1 of 5, going into three slots. That is a 3 x 2 x 1 = 6 permutation.

Then we have 5 ways to choose 1 of five. So there are five ways to do the 3 x 2 x 1 = 6 permutation.

So we have 6 x 5 = 30 permutations that include Drew and Eileen.

210 - 30 = 180

The correct answer is D.

An alternative way to handle it is to add up all the permutations that don't include both Drew and Eileen.

We have the ones with just Drew and the five others.

Drew can be in one of three slots.

D - -
- D -
- - D

Then the other 5 get arranged in the other two slots.

D 5 4

5 D 4

5 4 D

So we have 3 x 5 x 4 = 60 that include Drew but not Eileen.

The same thing can be done with Eileen and not Drew, for another 60.

Then, using just the others there are 5 x 4 x 3 = 60.

60 + 60 + 60 = 180

The correct answer is D.
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by 800_or_bust » Thu Jun 16, 2016 7:45 am
diegocml wrote:I haven't actually seen permutation in my GMAT studies yet. Having said that:

n!/(n-k)!

The fact that Drew and Eileen cannot serve together in any capacity makes me think that n = 6, but if I plug that I get > 6!/(6-3)! = 120, which is not even in the answer choices. Alternatively, if n = 7! I have 210, which I doubt is the correct answer.

I'm clueless, but I think the catch is with Drew and Eileen. Curious to see how to hack this problem.
Marty's solutions are both valid. Note that this is, indeed, a permutation and not a combination because order of the selections matters. By the fundamental counting principle, the total number of possible arrangements (without restrictions) is 210. But here I've given the restriction that Drew and Eileen cannot serve on the board together, so we need to determine all of the possible arrangements that violate this restriction and subtract that from the total.

There are six different cases which violate the restrictions, each of which can be accomplished in 5 different ways. Let's consider these from the perspective of Drew...

Case 1: Drew is President, Eileen is Treasurer, One of the remaining Five is Secretary

D E 5

Case 2: Drew is President, One of the Remaining Five is Treasurer, Eileen is Secretary

D 5 5

Case 3: Drew is Treasurer, One of the Remaining Five is Secretary, Eileen is President

E D 5

Case 4: Drew is Treasurer, One of the Remaining Five is President, Eileen is Secretary

5 D E

Case 5: Drew is Secretary, One of the Remaining Five is Treasurer, Eileen is President

E 5 D

Case 6: Drew is Secretary, One of the Remaining Five is President, Eileen is Treasurer

5 E D

Thus, there are 30 possible cases in which the restriction is violated. And hence the number of cases in which the restriction is adhered to is [spoiler]210 - 30 = 180[/spoiler]. Answer choice D.
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by 800_or_bust » Thu Jun 16, 2016 7:49 am
800_or_bust wrote:
diegocml wrote:I haven't actually seen permutation in my GMAT studies yet. Having said that:

n!/(n-k)!

The fact that Drew and Eileen cannot serve together in any capacity makes me think that n = 6, but if I plug that I get > 6!/(6-3)! = 120, which is not even in the answer choices. Alternatively, if n = 7! I have 210, which I doubt is the correct answer.

I'm clueless, but I think the catch is with Drew and Eileen. Curious to see how to hack this problem.
Marty's solutions are both valid. Note that this is, indeed, a permutation and not a combination because order of the selections matters. By the fundamental counting principle, the total number of possible arrangements (without restrictions) is 210. But here I've given the restriction that Drew and Eileen cannot serve on the board together, so we need to determine all of the possible arrangements that violate this restriction and subtract that from the total.

There are six different cases which violate the restrictions, each of which can be accomplished in 5 different ways. Let's consider these from the perspective of Drew...

Case 1: Drew is President, Eileen is Treasurer, One of the remaining Five is Secretary

D E 5

Case 2: Drew is President, One of the Remaining Five is Treasurer, Eileen is Secretary

D 5 5

Case 3: Drew is Treasurer, One of the Remaining Five is Secretary, Eileen is President

E D 5

Case 4: Drew is Treasurer, One of the Remaining Five is President, Eileen is Secretary

5 D E

Case 5: Drew is Secretary, One of the Remaining Five is Treasurer, Eileen is President

E 5 D

Case 6: Drew is Secretary, One of the Remaining Five is President, Eileen is Treasurer

5 E D

Thus, there are 30 possible cases in which the restriction is violated. And hence the number of cases in which the restriction is adhered to is [spoiler]210 - 30 = 180[/spoiler]. Answer choice D.
Alternatively, you can arrive at the correct answer using probabilities since the selections are made at random. The total number of arrangements (ignoring the restriction) is again 7 x 6 x 5 = 210.

Now determine the probability that both Drew and Eileen are selected to the board. For Drew the probability is 3/7 - there are three openings in the board and seven people who can fill the openings. And the probability that Eileen is also selected to the board, given that Drew has been selected, is 2/6 = 1/3. That is, there are now two openings on the board and six people who can fill the remaining two openings. The probability of both of these occurring is 3/7 x 1/3 = 1/7. Hence, in exactly 1 out of 7 of the total cases, both Drew and Eileen will be selected. 1/7th of 210 is 30. Subtract this figure from the total number of cases and the answer is again 180.
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by GMATGuruNY » Thu Jun 16, 2016 8:26 am
800_or_bust wrote:Just a question I came up. Figured I'd post it for my fellow would-be GMAT takers to attempt. Solution posted in 2 days.

Seven people - Anna, Bill, Corey, Drew, Eileen, Frank, and George - are being considered for a corporate board of officers, consisting of a president, a treasurer, and a secretary. In how many different ways can this board of officers be arranged if Drew and Eileen cannot serve together in any capacity?
Case 1: The board does not include Drew or Eileen
From the 5 remaining people, the number of ways to choose 3 to serve on the board = 5C3 = 10.

Case 2: The board includes Drew or Eileen but not both
From the 5 remaining people, the number of ways to choose 2 to serve with Drew or Eileen = 5C2 = 10.
Since there are 2 options for the third position -- either Drew or Eileen -- we multiply by 2:
10*2 = 20.

Case 1 + Case 2 = 10+20 = 30 cases.
In each case, there are 3 positions on the board: president, treasurer, secretary.
Number of ways to arrange these 3 positions = 3! = 6.
Multiplying the 30 cases by the number of ways each case can be arranged, we get:
30*6 = 180.

The correct answer is D.
Last edited by GMATGuruNY on Thu Jun 16, 2016 8:29 am, edited 1 time in total.
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by DavidG@VeritasPrep » Thu Jun 16, 2016 8:26 am
See here for a similar-ish official question involving probability: https://www.beatthegmat.com/probability- ... 72268.html
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by Brent@GMATPrepNow » Thu Jun 16, 2016 9:42 am
diegocml wrote:I haven't actually seen permutation in my GMAT studies yet. Having said that:
That's a great (i.e., true) statement. In fact, I'd say that one doesn't need to know the permutation formula if one knows how to apply the Fundamental Counting Principle.

Here's an article about this idea: https://www.gmatprepnow.com/articles/co ... %93-part-i

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by Matt@VeritasPrep » Thu Jun 23, 2016 4:41 pm
I'd be careful with the wording in the prompt, though. "How many ways can this board of officers be arranged" *could* be interpreted as "how many ways can you arrange the officers once you've chosen them", which is rather different from your intent.
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