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combination problem

by aleph777 » Sat Jul 31, 2010 5:08 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?

1) 2
2) 3
3) 4
4) 6
5) 12

I always resort to manually counting out possible arrangements with a little diagram, because I'm uncertain about solving this. With a simpler problem like this, it's not so bad, but for a larger problem, I'm sure you can't keep track with a drawing!

Can someone share the mathematical method for this? Is it 1(4) + 1(2), where 1 represents each individual student, and 4 and 2 represent the number of possible spaces available?

Thanks!
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by sanju09 » Sat Jul 31, 2010 5:22 am
aleph777 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?

1) 2
2) 3
3) 4
4) 6
5) 12

I always resort to manually counting out possible arrangements with a little diagram, because I'm uncertain about solving this. With a simpler problem like this, it's not so bad, but for a larger problem, I'm sure you can't keep track with a drawing!

Can someone share the mathematical method for this? Is it 1(4) + 1(2), where 1 represents each individual student, and 4 and 2 represent the number of possible spaces available?

Thanks!
Always resort to manually counting out possible arrangements when numbers are small as in this case.

When 2 students are to be seated in a row of 4 desks, we cannot keep more than 2 desks empty.

Following are the many ways of keeping 2 desks empty between the two students

A Ø Ø B or B Ø Ø A (= 2)

And following are the many ways of keeping 1 desk empty between the two students

A Ø B Ø or Ø A Ø B times 2 (= 4)

Total different ways = [spoiler]2 + 4 = 6.

D in our terms[/spoiler]
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by GMATGuruNY » Sat Jul 31, 2010 6:16 am
aleph777 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?

1) 2
2) 3
3) 4
4) 6
5) 12

I always resort to manually counting out possible arrangements with a little diagram, because I'm uncertain about solving this. With a simpler problem like this, it's not so bad, but for a larger problem, I'm sure you can't keep track with a drawing!

Can someone share the mathematical method for this? Is it 1(4) + 1(2), where 1 represents each individual student, and 4 and 2 represent the number of possible spaces available?

Thanks!
Good = Total - Bad

Total = total ways to arrange the 2 students:
1st student can choose from 4 desks
2nd student can choose from 3 desks
Total = 4*3 = 12.

Bad = arrangements that put the two students adjacent to each other.
Number of adjacent desk pairs = (desks 1 and 2) + (desks 2 and 3) + (desks 3 and 4) = 3.
Since the order of the 2 students can be reversed, 2*3 = 6 bad arrangements.

So Good = Total - Bad = 12 - 6 = 6.

The correct answer is D.
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