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Another one in Perm & Comb

Expert replies
by vabhs192003 » Fri Dec 28, 2012 2:04 am
Q. In how many ways can 3 letters out of five distinct letters A,B,C,D & E be arranged in a straight line so that A and B are never together?
1) 72
2) 42
3) 32
4) 52
5) 20

OA: 2

I can't get the hang of counting. The approach I followed is:
Lets say there are _ _ _ _ _ space where we have to put the letters. Taking different cases based on the position of A, we have:
case when 'A' is at 1st position: A _ _ _ _ ,'B' can be placed at 3rd, 4th, 5th position(i.e. 3 available places). Hence 3 x 3! ( 3! for the remaining letters).

Similarly for case when A is at 2nd position. _ A _ _ _, B has 2 places.Hence calculation is: 2 x 3!.
Similarly for case when A is at 3rd position: _ _ A _ _, calculation goes as: 2 x 3!.
Similarly for case when A is at 4th position: _ _ _ A _ , calculation goes as: 2 x 3!.
And finally when A is at 5th position: _ _ _ _ A, calculation goes as: 3 x 3!.
Adding all I am getting 72 as answer which is not OA. So what am I missing. Is my approach wrong?
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Source: — Problem Solving |

by Anurag@Gurome » Fri Dec 28, 2012 2:34 am
vabhs192003 wrote:Q. In how many ways can 3 letters out of five distinct letters A,B,C,D & E be arranged in a straight line so that A and B are never together?
The question is asking us to arrange three out of five letters in a straight line, not to arrange all the five letters.

Also the question has some ambiguity in the word "together". If "never together" means A and B will not be placed side by side, then the solution is as follows...

Number of ways to select 3 letters from 5 = 5C3 = 10
Number of ways to arrange three letters in a straight line = 3! = 6
Hence, number of possible arrangements without any restriction = 6*10 = 60

Now, A and B will be together only if they are among the selected three letters.
Number of ways to select 3 letters from 5 such that A and B are always selected = Number of ways to select 1 letter from 3 = 3C1 = 3
Now, in each of the above selection, we can arrange the three letters in 3! = 6 ways.
However, A and B will not be together only if they are placed on the first and last place, which can be done in 2 ways. Hence, A and B will be together for the rest (6 - 2) = 4 ways.
Hence, A and B will be always together for 3*4 = 12 arrangements

Hence, A and B will never be together for (60 - 12) = 48 arrangements


However, if "never together" means A and B will not be selected together, then the answer will be (60 - 18) = 42
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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by vabhs192003 » Fri Dec 28, 2012 10:22 am
Gee that was confusing. But I got a hang of your approach.

Thanks,
V
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