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probabilty

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probabilty

by xcusemeplz2009 » Mon Oct 26, 2009 8:06 am
one single person and two couples are to be seated at random in a row of five chairs.what is the probabilty that neither of the couples sits togathet in adjacent chairs

A)1/5
B)1/4
C)3/8
D)2/5
E)1/2

need short way to solve...
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by caspermonday » Mon Oct 26, 2009 10:37 am
by saying "neither of the couples sit together" do you mean that neither of two people making a couple sit together?
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by xcusemeplz2009 » Mon Oct 26, 2009 10:58 am
caspermonday wrote:by saying "neither of the couples sit together" do you mean that neither of two people making a couple sit together?
it means after the arrangement the partners are not seated together
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by kvamsy » Mon Oct 26, 2009 11:05 am
From my point of view simple way of solving this type ofquestions is to make a pictorial representation.

- - - - -

there are 5 total ways as a denominator,
Coming to the numerator there is only one possibility for the given condition that is

AA X BB , so answer could be 1/5 i guess.

Please let me know if this is correct
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by xcusemeplz2009 » Mon Oct 26, 2009 11:11 am
kvamsy wrote:From my point of view simple way of solving this type ofquestions is to make a pictorial representation.

- - - - -

there are 5 total ways as a denominator,
Coming to the numerator there is only one possibility for the given condition that is

AA X BB , so answer could be 1/5 i guess.

Please let me know if this is correct
you r getting the q wrong

we need a sort of arrangement like this
abxba or xabab or axbab.....like this i was able to find the ans but its time taking minimum time req is >3 min.... need a formula to solve....
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by sanjana » Tue Oct 27, 2009 7:44 am
IMO : 1/5

Whats the OA?
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by xcusemeplz2009 » Tue Oct 27, 2009 8:07 am
OA is 2/5
let the couple be aa,bb and the person be x

first case
fix X at position no. 1 then the adjacent seat will have 4 choices i.e 4 ways, let it be a
xa_ _ _

now the next seat has 2 choices(bb)....

xab_ _
and now next two seat will have 1 choice each

so tot choice in this case is 1*4*2*1*1=8

same no of choices will be true when x takes the last seat i.e
ababx
till here tot 8+8=16

case three..
when x is in second position...
_x_ _ _

for first position 4 choices ax_ _ _
now the third postion (i.e is next to x) will have two choice from bb in order to satisfy the conition
axbab
so 4*2*1*1*1=8

similarly for abaxb eight ways

total=16+8+8=32

case five when x is in centre
_ _ x _ _

1st seat has 4 choices ; second seat has 2 choices ;4th seat has 2 choices ; last 1 choice

4*2*2*2*1=16

tot=48

probability=48/120=2/5

but this is a layman soln....and consuming too much time...

can any body provide a more better soln...???
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by hongwang9703 » Tue Oct 27, 2009 6:27 pm
christ, even if i knew how to solve this question on the actual gmat, it would be very time consuming and not worthy of solving because my other questions will suffer more strict of time contraint
i got utterly defeated by the gmat.
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