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There are x children and y chairs arranged in a circle in a

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There are x children and y chairs arranged in a circle in a

by RBBmba@2014 » Wed Apr 22, 2015 6:34 am
There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

(1) x + y = 12

(2) There are more chairs than children.
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by RBBmba@2014 » Sat Apr 25, 2015 9:21 am
Math experts - could you please share your explanation for this problem ?

@Mitch/Brent/Dave/Rich - look forward to hearing from you!
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by [email protected] » Sat Apr 25, 2015 11:08 am
Hi RBBmba@2014,

This question is worded in a way that is open to interpretative bias. I think I know what the "intent" of the original author is, but the Official GMAT would not word this type of question in this way.

We're told that there are X children and Y chairs arranged in a circle and that X and Y are PRIMES. We're asked how many ways the X children can be seated in the Y chairs.

Fact 1: X + Y = 12

Since X and Y are PRIMES, the two values MUST be 5 and 7. At first glance, this might appear insufficient (since we don't know which number is which variable), but we haven't technically answered the question yet, so let's do a little more work and be sure....

IF...
X = 5
Y = 7

This means that there are 2 empty chairs, so we'd first have to choose 5 chairs to sit in (which is 7C5 = 21 different sets of 5 chairs) and then arrange the 5 children in those 5 chairs (which is 5! = 120 different arrangements. Under these circumstances, there are (21)(120) different arrangements.

IF....
X = 7
Y = 5

This means that there are 2 more children than seats, so 2 of the children won't be seated. The number of arrangements would be (7)(6)(5)(4)(3). While you might not immediately realize it, this is the exact SAME ANSWER as the other option. Here's why...

(21)(120) = (3)(7)(5)(4)(3)(2)(1) = (7)(2x3)(5)(4)(3) = (7)(6)(5)(4)(3)

So, either option yields the SAME RESULT.

Fact 1 is SUFFICIENT

Fact 2: There are more chairs than children

This gives us no numbers to work with, so there's no way to calculate anything.
Fact 2 is INSUFFICIENT

Final Answer: A

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by RBBmba@2014 » Sun Apr 26, 2015 9:40 pm
[email protected] wrote:Hi RBBmba@2014,

This question is worded in a way that is open to interpretative bias. I think I know what the "intent" of the original author is, but the Official GMAT would not word this type of question in this way.

We're told that there are X children and Y chairs arranged in a circle and that X and Y are PRIMES. We're asked how many ways the X children can be seated in the Y chairs.

Fact 1: X + Y = 12

Since X and Y are PRIMES, the two values MUST be 5 and 7. At first glance, this might appear insufficient (since we don't know which number is which variable), but we haven't technically answered the question yet, so let's do a little more work and be sure....

IF...
X = 5
Y = 7

This means that there are 2 empty chairs, so we'd first have to choose 5 chairs to sit in (which is 7C5 = 21 different sets of 5 chairs) and then arrange the 5 children in those 5 chairs (which is 5! = 120 different arrangements. Under these circumstances, there are (21)(120) different arrangements.

IF....
X = 7
Y = 5

This means that there are 2 more children than seats, so 2 of the children won't be seated. The number of arrangements would be (7)(6)(5)(4)(3). While you might not immediately realize it, this is the exact SAME ANSWER as the other option. Here's why...

(21)(120) = (3)(7)(5)(4)(3)(2)(1) = (7)(2x3)(5)(4)(3) = (7)(6)(5)(4)(3)

So, either option yields the SAME RESULT.

Fact 1 is SUFFICIENT

Fact 2: There are more chairs than children

This gives us no numbers to work with, so there's no way to calculate anything.
Fact 2 is INSUFFICIENT

Final Answer: A

GMAT assassins aren't born, they're made,
Rich
Hi Rich,
Thanks for your reply.

I got this one BUT my SPECIFIC DOUBT was isn't this solution good for linear arrangement only ?

I mean, for circular arrangement, I thought solution would be 7C5*4! because after selecting 5 chairs from 7 (or 5 persons from 7), 5 persons could be arranged in a circle in (5-1)!=4! ways. Right ?

Correct me please if wrong!

P.S: IRRESPECTIVE of the solution, although oa will remain as A
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by RBBmba@2014 » Wed Apr 29, 2015 2:01 am
Rich - could you please share your thoughts on my above post ?

Thank you!
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by [email protected] » Wed Apr 29, 2015 9:44 am
HI RBBmba@2014,

When I mentioned in an earlier post that the prompt was open to interpretive bias, this is exactly what I was referring to. You bring up a good point; one that the author of this question doesn't seem to address (there's NO reference to what constitutes a "unique arrangement"). As such, I have to go with how the question is written and do the math accordingly.

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by RBBmba@2014 » Wed Apr 29, 2015 10:06 am
[email protected] wrote:HI RBBmba@2014,

When I mentioned in an earlier post that the prompt was open to interpretive bias, this is exactly what I was referring to. You bring up a good point; one that the author of this question doesn't seem to address (there's NO reference to what constitutes a "unique arrangement"). As such, I have to go with how the question is written and do the math accordingly.

GMAT assassins aren't born, they're made,
Rich
Rich - I think, I didn't get you here because it's given in the Qs that it's circular arrangement. Then can you please let me know why we won't consider the method of circular arrangement ?
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by Matt@VeritasPrep » Mon May 04, 2015 12:08 am
Yeah, you need to consider circular arrangements, not linear ones.

For instance, if we have x = 7 and y = 5, we would first choose the 5 kids who actually get seats, which is (7 choose 5), or 21. From there, we'd arrange our five kids in a circle, which we can do in 4! ways, for a total of 21 * 24 = 504 arrangements.

If x = 5 and y = 7, however, we're in trouble. As with any circular arrangement, we need to fix one of the kids, then arrange everything in terms of that kid. This means that we're arranging four kids and two blank seats, or ABCDXX. This gives 6! / 2! = 360 arrangements, which is NOT THE SAME ANSWER.

So we need to know which case we're dealing with, and the answer is C.
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