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vabhs192003
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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?
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?
