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by becnil » Sun Feb 07, 2010 1:39 pm
How many different ways can 2 students be seated in a row of 4 desks so that there is always at least one empty desk between the students?

I will give the answer a little bit later..
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by shashank.ism » Sun Feb 07, 2010 1:39 pm
becnil wrote:How many different ways can 2 students be seated in a row of 4 desks so that there is always at least one empty desk between the students?

I will give the answer a little bit later..
The correct answer is 3
Let us denote desks by | and arrange the bench below.
| | | | Now we have to place 2 students be placed such that at least one empty desk between the students.

so following arrangement can be made for following condition boy seats at(1,3),(1,4),(2,4)
hence total no. =3
Last edited by shashank.ism on Sun Feb 07, 2010 1:46 pm, edited 2 times in total.
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by becnil » Sun Feb 07, 2010 1:41 pm
Sorry, its not 22
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by becnil » Sun Feb 07, 2010 1:42 pm
I picked up 3 at first and I was wrong !!
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by harsh.champ » Sun Feb 07, 2010 1:45 pm
becnil wrote:How many different ways can 2 students be seated in a row of 4 desks so that there is always at least one empty desk between the students?

I will give the answer a little bit later..
I guess 6 only.
If a student is on the 1st desk,the other student can be on the 3rd or 4th desk.(2ways)
If a student is on the 2nd desk,the other student can be on the 4th desk only. (1 way)
Selection of the student can be done in 2 ways,(either S1 or S2)
So, answer will be 6 i guess.
If you are not considering the permutations it would be 3.
Hey becnil plz post the answer choices as well as the solution.
Last edited by harsh.champ on Sun Feb 07, 2010 1:50 pm, edited 3 times in total.
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by shashank.ism » Sun Feb 07, 2010 1:48 pm
becnil wrote:I picked up 3 at first and I was wrong !!
Ok harsh you are correct the two boys can also be arranged at these 3 places in 2 ways ..so total no. = 2x3 =6.
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by becnil » Sun Feb 07, 2010 1:56 pm
The correct answer is 6. I picked up 3, thinking that the order did not matter in case of students (X sitting next to Y is same as Y sitting next to X). But the source (Kaplan) considered the order and the answer is 6.
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by harsh.champ » Sun Feb 07, 2010 2:02 pm
becnil wrote:The correct answer is 6. I picked up 3, thinking that the order did not matter in case of students (X sitting next to Y is same as Y sitting next to X). But the source (Kaplan) considered the order and the answer is 6.
Well,I guess then my answer was correct.
becnl,can you post the 5 answer choices also.

And was my problem approach correct or did kaplan use some other technique?
Regarding the question,I think it could have been phrased in a better manner.Over here,it is doubtful whether to consider the order(i.e. right to left )or not .
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by shashank.ism » Sun Feb 07, 2010 7:37 pm
becnil wrote:The correct answer is 6. I picked up 3, thinking that the order did not matter in case of students (X sitting next to Y is same as Y sitting next to X). But the source (Kaplan) considered the order and the answer is 6.
Becnil Though In first impression I also picked up 3. But Order of students really does matter and If you change the order of students, you will find a different look.
For e.g. here there were 2 students only. But if there were in any case 3 or more students, then think once again, the order of students will really matter and it will give u different combinations.
So I think Kaplan consideration of order is legit and you must carry it on for any further problems of this type.
Harsh you are very correct in your approach and be sure to consider the order.
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by Stuart@KaplanGMAT » Mon Feb 08, 2010 12:06 am
becnil wrote:How many different ways can 2 students be seated in a row of 4 desks so that there is always at least one empty desk between the students?

I will give the answer a little bit later..
When we count arrangements, order is always relevant.

For this type of question, it's often helpful to name the objects; here, let's call our students A and B. Let's call the two empty seats X and X (since the empty seats are identical, we don't need to worry about swapping them to create new arrangements).

Since it's such a simple situation (4 spots for 2 people), it's almost certainly going to be fastest to just write out the options:

AXBX
AXXB
XAXB

we now realize that we can just swap A and B in their seats, to the total number of arrangements is 3*2 = 6.
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