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Manhattan GMAT 700+ problem - July 31

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Kevin Community Manager Default Avatar
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Manhattan GMAT 700+ problem - July 31

Post Mon Jul 31, 2006 12:07 pm
Most of our students Manhattan GMAT are trying to break the 700+ barrier. As a result, we've developed our own math problems written at the 700+ level; these are the types of questions you'll WANT to see, when you are working at that level. Try to solve this 700+ level problem (I'll post the solution next Monday).

Question (Prime Seats)
There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?

(1) x + y = 12
(2) There are more chairs than people.


(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

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Kevin Fitzgerald
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800-576-4626

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Kevin Community Manager Default Avatar
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Post Mon Aug 07, 2006 8:19 am
Answer
This question is simply asking us to come up with the number of permutations that can be formed when x people are seated in y chairs. It would seem that all we require is the values of x and y. Let's keep in mind that the question stem adds that x and y must be prime integers.

(1) SUFFICIENT: If x and y are prime numbers and add up to 12, x and y must be either 7 and 5 or 5 and 7. Would the number of permutations be the same for both sets of values?

Let's start with x = 7, y = 5. The number of ways to seat 7 people in 5 positions (chairs) is 7!/2!. We divide by 2! because 2 of the people are not selected in each seating arrangement and the order among those two people is therefore not significant. An anagram grid for this permutation would look like this:

A B C D E F G
1 2 3 4 5 N N

But what if x = 5 and y = 7? How many ways are there to position five people in 7 chairs? It turns out the number of permutations is the same. One way to think of this is to consider that in addition to the five people (A,B,C,D,E), you are seating two ghosts (X,X). The number of ways to seat A,B,C,D,E,X,X would be 7!/2!. We divide by 2! to eliminate order from the identical X's.

Another way to look at this is by focusing on the chairs as the pool from which you are choosing. It's as if we are fixing the people in place and counting the number of ways that different chair positions can be assigned to those people. The same anagram grid as above would apply, but now the letters would correspond to the 7 chairs being assigned to each of the five fixed people. Two of the chairs would be unassigned, and thus we still divide by 2! to eliminate order between those two chairs.

(2) INSUFFICIENT: This statement does not tell us anything about the values of x and y, other than y > x. The temptation in this problem is to think that you need statement 2 in conjunction with statement 1 to distinguish between the x = 5, y= 7 and the x = 7, y = 5 scenarios.

The correct answer is A: Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not

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Kevin Fitzgerald
Director of Marketing and Student Relations
Manhattan GMAT
800-576-4626

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dblazquez Senior | Next Rank: 100 Posts
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Post Mon Aug 07, 2006 6:17 am
gosh you're right, it's the first quant flash card Embarassed

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jtsgmat Newbie | Next Rank: 10 Posts Default Avatar
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Post Mon Aug 07, 2006 3:08 am
I would go with (A).

There are 2 possible ways to combine X and Y to be equal to 12 using positive prime numbers.

7 + 5 = 12
5 + 7 = 12

Now, if there are only 5 chairs but 7 people then the arrangmements can be viewed as follows(let C = chair):

C1 C2 C3 C4 C5
# Options 7 6 5 4 3

Total arragements = 2 520

If there are 7 chairs and 5 people, the we can use the permutation formula (since order matters):

7! / (7-5)! = (7*6*5*4*3*2!)/2! = 7*6*5*4*3 = 2 520

Therefore, it does not matter if there are 5 chairs and 7 people or 7 chairs and 5 people becuase the total number of arrangments will be equivalent.

Hope this is correct.

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dblazquez Senior | Next Rank: 100 Posts
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Post Mon Aug 07, 2006 5:47 am
nice explanation, butwhy 11 and 1 is not a valid option? is not 1 a valid prime number?

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jtsgmat Newbie | Next Rank: 10 Posts Default Avatar
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Post Mon Aug 07, 2006 6:10 am
1 is not a prime number.

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