probability

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2008: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Tue Apr 08, 2008 10:09 am; Thanked: 4 times

by 2008 » Sat Aug 09, 2008 8:25 am

two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a.1/5
b.1/4
c.3/8
d.2/5
e.1/2

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Source: Beat The GMAT — Problem Solving | Sat Aug 09, 2008 8:25 am

2008: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Tue Apr 08, 2008 10:09 am; Thanked: 4 times

by 2008 » Sat Aug 09, 2008 8:45 am

OA is D

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Sat Aug 09, 2008 9:02 am

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..

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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

by raunekk » Sat Aug 09, 2008 9:17 am

i cant give u a better explanation than Ian

@sudhir

we all need a quicker method

https://www.beatthegmat.com/war-of-the-r ... 12457.html

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Sat Aug 09, 2008 9:29 am

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...

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mmgmat2008: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Tue Nov 11, 2008 6:46 am

don't understand

by mmgmat2008 » Thu Nov 27, 2008 4:03 pm

sudhir3127 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..

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
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: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

by logitech » Thu Nov 27, 2008 4:17 pm

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!

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Thu Nov 27, 2008 4:38 pm

Assume C takes the 2 second place its (4*1*2*1*1) = 8 ways

Sudhir,
Why is this not

4*1*3*1*1 = 12??

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Thu Nov 27, 2008 4:41 pm

I got it sorry!

If its a,B,b left it has to be B or b for the 3rd spot since a has to come between them!!!

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