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...
probabilty
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It does not matter how many times you get knocked down , but how many times you get up
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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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it means after the arrangement the partners are not seated togethercaspermonday wrote:by saying "neither of the couples sit together" do you mean that neither of two people making a couple sit together?
It does not matter how many times you get knocked down , but how many times you get up
From my point of view simple way of solving this type ofquestions is to make a pictorial representation.
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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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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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you r getting the q wrongkvamsy wrote:From my point of view simple way of solving this type ofquestions is to make a pictorial representation.
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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
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....
It does not matter how many times you get knocked down , but how many times you get up
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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...???
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...???
It does not matter how many times you get knocked down , but how many times you get up
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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.