arranging problem

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arranging problem

by pappueshwar » Wed Feb 22, 2012 9:21 am
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?[spoiler] i feel the answer is 4 , is that right ?[/spoiler]

(A) 2
(B) 3
(C) 4
(D) 6
(E) 12
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by rijul007 » Wed Feb 22, 2012 10:00 am
pappueshwar 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?[spoiler] i feel the answer is 4 , is that right ?[/spoiler]

(A) 2
(B) 3
(C) 4
(D) 6
(E) 12
Atleast one desk should be empty means one or more desks should be vacant

Total no of arrangements = 4C2 * 2 = 12
No of arrangement when the students are together = 3*2 = 6
No of arrangements with atleast one empty desk = 12 - 6 = 6

Option D





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by GMATGuruNY » Wed Feb 22, 2012 10:29 am
pappueshwar 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?[spoiler] i feel the answer is 4 , is that right ?[/spoiler]

(A) 2
(B) 3
(C) 4
(D) 6
(E) 12
Desks:
The students can occupy the following pairs of desks: 1 and 3, 1 and 4, 2 and 4.
Number of options = 3.

Students:
The number of ways to arrange the 2 students = 2! = 2.

To combine the options above, we multiply:
3*2 = 6.

The correct answer is D.

We should recognize that the answer choices are SMALL, implying that we can quickly WRITE OUT all of the possible arrangements.

Here are all of the ways to arrange two students A and B and two empty desks E and E so that A and B are not adjacent:
AEBE
AEEB
EAEB
BEAE
BEEA
EBEA
Total ways = 6.
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