probability
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sudhir3127
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My Answer is 2/5
Let the couple be Aa Bb and C( single person )
out of 5 places
Assume if C takes the 1st place
A can take the first place in 4 ways , a cannot take the second place. it has to be taken by wither B or b hence is 2 ways and the last 2 places in 1 way
hence 1*4*2*1*1 = 8 ways
Assume C takes the 5 th place - 8 ways same as above.
Assume C takes the 2 second place its (4*1*2*1*1) = 8 ways
Assume C takes the 4th place - 8 ways same as above
If C takes the 3rd place
1st place in 4 ways , 2nd place in 2 ways , 4th place in 2 place and 5th place in 1 way
hence its 4*2*1*2*1 = 16.
therefore its
8+8+8+8+16 = 48 ways.
5 people can arrange themselves in 5! = 120
thus the probability is 48/120 which is 2/5.
hope it helps..
Let the couple be Aa Bb and C( single person )
out of 5 places
Assume if C takes the 1st place
A can take the first place in 4 ways , a cannot take the second place. it has to be taken by wither B or b hence is 2 ways and the last 2 places in 1 way
hence 1*4*2*1*1 = 8 ways
Assume C takes the 5 th place - 8 ways same as above.
Assume C takes the 2 second place its (4*1*2*1*1) = 8 ways
Assume C takes the 4th place - 8 ways same as above
If C takes the 3rd place
1st place in 4 ways , 2nd place in 2 ways , 4th place in 2 place and 5th place in 1 way
hence its 4*2*1*2*1 = 16.
therefore its
8+8+8+8+16 = 48 ways.
5 people can arrange themselves in 5! = 120
thus the probability is 48/120 which is 2/5.
hope it helps..
i cant give u a better explanation than Ian 
@sudhir
we all need a quicker method
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@sudhir
we all need a quicker method
https://www.beatthegmat.com/war-of-the-r ... 12457.html
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sudhir3127
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heres another approach
total ways is 5!
one couple sits together, hence the number of ways is 2! x 4!
the no. of ways other couple can sit is 2! x 4!
total = 2 x 2! x 4! ways
But in above cases include both couples are together twice.
hence we need to subtract one.
The number of ways both couples sit together is 3! x 2! x 2!.
Probability that any of the couples sitting in adjacent chairs =
= [(2 x 2! x 4! ) - (3! x 2! x 2!)]/5! =3/5
neither of the couples sit together in adjacent chairs = 1 - 3/5 = 2/5
Hope its better raunekk and faster as well...
total ways is 5!
one couple sits together, hence the number of ways is 2! x 4!
the no. of ways other couple can sit is 2! x 4!
total = 2 x 2! x 4! ways
But in above cases include both couples are together twice.
hence we need to subtract one.
The number of ways both couples sit together is 3! x 2! x 2!.
Probability that any of the couples sitting in adjacent chairs =
= [(2 x 2! x 4! ) - (3! x 2! x 2!)]/5! =3/5
neither of the couples sit together in adjacent chairs = 1 - 3/5 = 2/5
Hope its better raunekk and faster as well...
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mmgmat2008
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A can take the first place in 4 ways , a cannot take the second place. it has to be taken by wither B or b hence is 2 ways and the last 2 places in 1 waysudhir3127 wrote:My Answer is 2/5
Let the couple be Aa Bb and C( single person )
out of 5 places
Assume if C takes the 1st place
A can take the first place in 4 ways , a cannot take the second place. it has to be taken by wither B or b hence is 2 ways and the last 2 places in 1 way
hence 1*4*2*1*1 = 8 ways
Assume C takes the 5 th place - 8 ways same as above.
Assume C takes the 2 second place its (4*1*2*1*1) = 8 ways
Assume C takes the 4th place - 8 ways same as above
If C takes the 3rd place
1st place in 4 ways , 2nd place in 2 ways , 4th place in 2 place and 5th place in 1 way
hence its 4*2*1*2*1 = 16.
therefore its
8+8+8+8+16 = 48 ways.
5 people can arrange themselves in 5! = 120
thus the probability is 48/120 which is 2/5.
hope it helps..
hence 1*4*2*1*1 = 8 ways
Can you explain how you get 1*4*2*1*1= 8 ways? I have the A takes the first place, but come up with more than 4 ways. Please help me out.
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logitech
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Let me try:
Couples A1A2 B1B2 and single S
lets have the A1B2 A2B1 and S as three units
3x2x1=6 ways both couples can interchange so 2x2 = 4
4x6 =24
since we can also have A1B1 and A2B2 it is 24 too
48/5!
Couples A1A2 B1B2 and single S
lets have the A1B2 A2B1 and S as three units
3x2x1=6 ways both couples can interchange so 2x2 = 4
4x6 =24
since we can also have A1B1 and A2B2 it is 24 too
48/5!
LGTCH
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