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# 5 people, including A and B, are to be seated in a row. How

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5 people, including A and B, are to be seated in a row. How [#permalink]

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05 May 2007, 12:32
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5 people, including A and B, are to be seated in a row. How many different ways can they be seated subject to a constraint that A and B can not sit next to each other?

A) 120
B) 96
C) 72
D) 48
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05 May 2007, 14:03
answer = all permutations - constraint
all permutations = 5!
error here: constraint = 8 (ABxxx, BAxxx, xABxx, xBAxx, xxABx, xxBAx, xxxAB, xxxBA)
answer = 5! - 8 = 5*4*3*2 - 8 = 120 - 8 = 112

Last edited by kekc on 05 May 2007, 14:35, edited 1 time in total.
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05 May 2007, 14:07
kekc wrote:
answer = all permutations - constraint
all permutations = 5!
constraint = 8 (ABxxx, BAxxx, xABxx, xBAxx, xxABx, xxBAx, xxxAB, xxxBA)
answer = 5! - 8 = 5*4*3*2 - 8 = 120 - 8 = 112

constrain is much bigger
you can not assume that all other people are x
they are x,y and z
So only for your first option there will be:
ABxyz,ABxzy,ABzyx etc etc....
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05 May 2007, 14:17
E!
Can't prove me wrong there
but I'll also go with C

A&B never together = Total - A&B always together
Total =5! =120
A&B always together = 4!2! =48
(for A&B always together, treat A&B as one unit so you have total 4 units and hence the 4!, but A&B can switch places amongst themselves and so the 2!)

120 - 48 = 72
and that's both E and C
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05 May 2007, 14:19
yeah.... I forgot to post E
but it doesn't matter
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05 May 2007, 14:33
Caas, you are right, in my solution constraint should be multiplied by 6.
So, 5! - 6*8 = 120 - 48 = 72.
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05 May 2007, 14:47
It should be 'C'.

Total number of ways to seat 5 people in a row = 5! = 120
number of ways A and B will seat with each other = 4!2! = 48
So number of ways to seat 5 people when A and B don't sit next to each other = 120-48 =72
05 May 2007, 14:47
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