Problem Solving

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.

Hi,
From(1):
(x,y) can be (5,7) or (7,5)
If (x,y) is (5,7) -> number of ways is 7P5
If (x,y) is (7,5) -> number of ways is 0
Not Sufficient
From(2):
No info. about x and y
Sufficient
Both(1) and (2):
x<y. So, (x,y) is (5,7) -> number of ways is 7P5
Sufficient

Hence, C

Frankenstein wrote:Hi,
From(1):
(x,y) can be (5,7) or (7,5)
If (x,y) is (5,7) -> number of ways is 7P5
If (x,y) is (7,5) -> number of ways is 0
Not Sufficient
From(2):
No info. about x and y
Sufficient
Both(1) and (2):
x<y. So, (x,y) is (5,7) -> number of ways is 7P5
Sufficient

Hence, C

No Frank! Try an other one. This is really trickY!

Ozlemg wrote:

Frankenstein wrote:Hi,
From(1):
(x,y) can be (5,7) or (7,5)
If (x,y) is (5,7) -> number of ways is 7P5
If (x,y) is (7,5) -> number of ways is 0
Not Sufficient
From(2):
No info. about x and y
Sufficient
Both(1) and (2):
x<y. So, (x,y) is (5,7) -> number of ways is 7P5
Sufficient

Hence, C

No Frank! Try an other one. This is really trickY!

OMG..It's not C? Could you post OA,OE and source. I would be very interested.

SOLUTION WITH EXPLANATION :

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.

(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.


ps : i dont know the source!

Hi,
Thanks for posting the solution.
First, I seriously doubt the circulation of this question. I think this could have been scrapped.
Second, this defies the basic assumptions of Permutations and Combinations. When we say 5 people are to be seated in 7 chairs, we mean that all 5 have to be seated and hence we write 7P5. Similarly, when we say 7 people to seated in 5 chairs, how can we consider the case of only 5 of them being seated and the other two not being seated. If this logic were true, I would calculate the first case of seating 5 people in 7 chairs as a combination of many cases - 1 person seated and the other 6 standing, then 2 persons seated and 3 standing and so on. This simply doesn't sound great. Whatever be the intention of the question framer, I will not agree with this. I better ignore this question.

Btw, are you convinced with the solution you posted?

if 5 people can be seated in 7 chairs then if we use the counting way it would be..the 1st person can sit in any 7 chair, the 2nd person in any 6 chair the 3rd person in any 5 chair..7x6x5x4x3 should be the total ways in which it is possible...and incase 7 people are sitting in any 5 chairs it would be the 1st person can sit in any 5 chair, the 2nd person in any 4 chair...so number of ways would be 5x4x3x2x1....please advice where I am going wrong on this one..

I think I've seen a similar MGMAT problem of this nature. If you have excess chairs, you treat the unused chairs as being occupied by ghosts..lol

I agree with Frank here. This is not a good solution. If you want to know "How many ways is there to seat 7 people?" then, seven people better be seated. If they are not, it should not be counted as a way to seat seven people. Imagine if 1,000 people were coming to see a band in an auditorium that only had 3 chairs. And the band's agent says to the manager of the auditorium, "Ummm.....is there any way we can seat 1,000 people in 3 chairs?" And the manager says "Yes! of course! there are 997,002,000 ways to do it! 1000*999*998. Don't you know anything about math???"

Pretty ridiculous.

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.

Answer :

consider this way

From Statement (1)
only possible combination are

People, Chairs
7,5 <- case 1
5,7 <- case 2

Consider both case
Case 1>
we have 7 people and 5 chairs
7P5 or P(7,5) = 7!/(7-5)! = 7!/2!

case 2>
now we have 5 people and 7 chairs

consider this way

- - - - - - - (there are 7 chairs)

so first person has 7 options to choose from
second person has 6 options to choose from
third person has 5 option to choose from
fourth person has 4 option to choose from
fifth person has 3 option to choose from

7 * 6 * 5 * 4 * 3 = 7!/2! (you can calculate it)

so both option give the same output

Hence Official Answer is A.

Cheers,
nikes