permuatations and probability

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chaitanyareddy: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Fri Feb 27, 2009 12:45 am; Thanked: 1 times; GMAT Score:620

by chaitanyareddy » Sat Aug 07, 2010 2:41 am

Hi Experts,

I am facing real tough time with permutation and combinations and also probability. anyone of you , if you are having a good ebook or notes regarding P and C and probaility , please post the same..

Could you please explain me the solution for this problem.

Q. TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??

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Source: Beat The GMAT — Problem Solving | Sat Aug 07, 2010 2:41 am

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Sat Aug 07, 2010 6:16 am

chaitanyareddy wrote:Hi Experts,

I am facing real tough time with permutation and combinations and also probability. anyone of you , if you are having a good ebook or notes regarding P and C and probaility , please post the same..

Could you please explain me the solution for this problem.

Q. TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??

Total no. of possibilities = 5!

Case 1:
Let the single person is in the 1st seat.

2nd seat: Any one person from the remaining 4 persons, which can be done in 4 ways.
3rd seat can be filled in 2 ways as couples cannot be seated next to each other.
4th seat can be filled in 1 way, which can be occupied by the partner of the person in the 2nd seat.
5th seat can be filled in 1 way, which can be occupied by the partner of the person in the 3rd seat.

So, no. of ways of seating 2 couples and a single person when single person is in the 1st seat = 4*2*1*1 = 8 ways, which will also be the case when single person is seated in 5th seat.

Case 2:

Single person in the 2nd seat.
1st seat: Any one person from the 4 persons, which can be done in 4 ways.
3rd seat: 2 ways
4th seat: 1 way
5th seat: 1 way

So, no. of ways of seating 2 couples and a single person when single person is in the 2nd/4th seat = 4*2*1*1 = 8 ways, which will also be the case when single person is seated in 4th seat.

Case 3:
Single person in the 3rd seat.
1st seat: Any one person from the 4 persons, which can be done in 4 ways.
2nd seat: 2 ways (any person from the other couple)
4th seat: 2 way (partner of any of the 2 couples on 1st and 2nd seat)
5th seat: 1 way

So, no. of ways of seating 2 couples and a single person when single person is in the 3rd seat = 4*2*2*1 = 16 ways

Required probability = (16 + 16 + 16)/ 5! = 48/(5*4*3*2*1) = 48/120 = 2/5

The correct answer is 2/5.

Rahul Lakhani
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pnk: Master | Next Rank: 500 Posts; Posts: 292; Joined: Fri Jul 17, 2009 8:39 amFollowed by: 1 member; Thanked: 6 times

by pnk » Sat Aug 07, 2010 9:47 am

Hi Rahul,

We can also solve this by finding couples always sitting together and thn subtracting from 1.

Using this, could you find error in my approach. I am not able to find one:

5 people = individual 1 (I1) + couple 1 (C1)+couple 2 (C2) => 3 people only

I will try to find possibilities where c1 and c2 sit together. In that case, we have 3 persons sitting =3!. Plus each set of couple can arrange among themselves in 2! ways. Therefore nos of ways in which 5 people can sit (with couples sitting adjacent) = 3! 2! 2! = 24

Probability of not sitting together = (1 - 24/5!) = 4/5

Can someone help me where I am wrong

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Sat Aug 07, 2010 10:11 am

pnk wrote:Hi Rahul,

We can also solve this by finding couples always sitting together and thn subtracting from 1.

Using this, could you find error in my approach. I am not able to find one:

5 people = individual 1 (I1) + couple 1 (C1)+couple 2 (C2) => 3 people only

I will try to find possibilities where c1 and c2 sit together. In that case, we have 3 persons sitting =3!. Plus each set of couple can arrange among themselves in 2! ways. Therefore nos of ways in which 5 people can sit (with couples sitting adjacent) = 3! 2! 2! = 24

Probability of not sitting together = (1 - 24/5!) = 4/5

Can someone help me where I am wrong

You missed the probability where each of the couples can sit together.
Suppose, only first couple sits together, then probability is
3P2 * 2! * 2! = 24
Similar is the case for second couple sitting together
Hence 3P2 * 2! * 2! = 24

Total number of ways of arranging 5 people = 5! = 120 ways

From this 24+24+24 have to be subtracted.
Hence prob ability = 1-(72/120) = 2/5

Hope this helps!!

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neerajbansal: Senior | Next Rank: 100 Posts; Posts: 61; Joined: Fri Feb 19, 2010 10:12 am; Thanked: 3 times

by neerajbansal » Sat Aug 07, 2010 4:56 pm

Rahul,

What am I missing in my explanation ? Can you explain ?

Q. TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??

--> p(event) --- Desired Outcomes/Total possible

Total Possible --> 5!

Desired Outcomes --> Assume 4 couples are ties with a rope and they are collectively called CCCC. so they can sit with the Single guy (S) in 2! ways. and CCCC can sit together in 4! ways.

So 4!2! is the way 4 couples can sit together with S and they are all together.

So P(4 couples sitting next to each other) is 4!2!/5!

So P(4 couples NOT sitting next to each other) = 1- P(4 couples sitting next to each other)

1 - 2/5

3/5 ans

Or should I think of a couple as a single entity

With that in mind:

c1-c2-s

The number of ways the couples are always next to each other are : 3! * 2!* 2!

so P(that all couple are next to each other) -- > (3! * 2!* 2! )/5! --> 1/5

So P(4 couples NOT sitting next to each other) = 1- P(4 couples sitting next to each other)

1- 1/5

4/5 ans..

CAN YOU TELL ME WHERE AM I GOING WRONG

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Aug 08, 2010 4:11 am

chaitanyareddy wrote:Hi Experts,

I am facing real tough time with permutation and combinations and also probability. anyone of you , if you are having a good ebook or notes regarding P and C and probaility , please post the same..

Could you please explain me the solution for this problem.

Q. TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??

We could treat this as an overlapping groups problem:

Total = G(1) + G(2) - Both + Neither

Let's say we have couple AB, couple CD, and lonely person E.

Total arrangements = (arrangements with AB together) + (arrangements with CD together) - (arrangements with both AB and CD together) + (arrangements with neither AB nor CD together)

Total arrangements = 5! = 120
Total arrangements with AB together means we're arranging the 4 elements AB, C, D, and E = 4!*2 = 48 (We multiply by 2 since AB can be reversed to be BA.)
Total arrangements with CD together means we're arranging the 4 elements CD, A, B and E = 4!*2 = 48 (We multiply by 2 since CD can be reversed to be DC.)
Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3!(4) = 24 (We multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed.)

Plugging into the overlapping groups formula, we get:

120 = 48 + 48 - 24 + N
120 = 72 + N
N = 48

So P(neither couple sitting together) = 48/120 = 2/5.

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pnk: Master | Next Rank: 500 Posts; Posts: 292; Joined: Fri Jul 17, 2009 8:39 amFollowed by: 1 member; Thanked: 6 times

by pnk » Sun Aug 08, 2010 9:19 am

Hi Mitch,

I agree with your logic but why my approach gives me a different answer:

Probability of couple never being together = 1 - Probability of couple always together.

Couple AB, CD; Individual E

Arrangement with AB, CD together = 3! 2! 2! (3! bcoz AB, CD, E is 1 group each - and these 3 groups can be arranged among themselves in 3! ways; 2! each for AB and CD arranging among themelves) = 24

So, probability = 1 - 24/120 = 4/5. Could you pls explain where I am at fault.

GMATGuruNY wrote:
We could treat this as an overlapping groups problem:

Total = G(1) + G(2) - Both + Neither

Let's say we have couple AB, couple CD, and lonely person E.

Total arrangements = (arrangements with AB together) + (arrangements with CD together) - (arrangements with both AB and CD together) + (arrangements with neither AB nor CD together)

Total arrangements = 5! = 120
Total arrangements with AB together means we're arranging the 4 elements AB, C, D, and E = 4!*2 = 48 (We multiply by 2 since AB can be reversed to be BA.)
Total arrangements with CD together means we're arranging the 4 elements CD, A, B and E = 4!*2 = 48 (We multiply by 2 since CD can be reversed to be DC.)
Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3!(4) = 24 (We multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed.)

Plugging into the overlapping groups formula, we get:

120 = 48 + 48 - 24 + N
120 = 72 + N
N = 48

So P(neither couple sitting together) = 48/120 = 2/5.

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muralithe1: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Fri Mar 05, 2010 9:50 pm; Thanked: 1 times

by muralithe1 » Sun Aug 08, 2010 9:57 am

Hi PNK,
i did the same fault...Then realized with GMAT guru explanation..

Probability of couple never being together = 1 - Probability of couple always together.

Couple AB, CD; Individual E
---->U have forgotten something here. This as well includes AB,C,E,D means only one couple is treated as one unit and other not...But this scenario also shld be excluded as the problem says no couple shld sit together...

Arrangement with AB, CD together = 3! 2! 2! (3! bcoz AB, CD, E is 1 group each - and these 3 groups can be arranged among themselves in 3! ways; 2! each for AB and CD arranging among themelves) = 24

So, probability = 1 - 24/120 = 4/5. Could you pls explain where I am at fault.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Aug 08, 2010 11:32 am

pnk wrote:Hi Mitch,

I agree with your logic but why my approach gives me a different answer:

Probability of couple never being together = 1 - Probability of couple always together.

Couple AB, CD; Individual E

Arrangement with AB, CD together = 3! 2! 2! (3! bcoz AB, CD, E is 1 group each - and these 3 groups can be arranged among themselves in 3! ways; 2! each for AB and CD arranging among themelves) = 24

So, probability = 1 - 24/120 = 4/5. Could you pls explain where I am at fault.

You need to account for all the different ways that one or both couples could sit together. Here's an equation that you could use:

P(neither couple together) = 1 - P(only AB together) - P(only CD together) - P(both AB and CD together)

Number of ways to arrange AB, CD and E = 3! * 4 = 24.
So P(AB and CD together) = 24/120 = 1/5.

Number of ways to arrange AB, C, D and E with CD not together = (Total arrangements of AB, C, D, and E) - (number of arrangements with both AB and CD together) = (4! * 2) - 24 = 24.
So P(only AB together) = 24/120 = 1/5.

Number of ways to arrange CD, A, B and E with AB not together = (Total arrangements of CD, A, B, and E) - (number of arrangements with both AB and CD together) = (4! * 2) - 24 = 24
So P(only CD together) = 24/120 = 1/5.

Thus P(neither couple together) = 1 - 1/5 - 1/5 - 1/5 = 2/5.

Hope this helps!

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aarati: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Thu Aug 05, 2010 3:15 am

by aarati » Sun Aug 08, 2010 11:53 pm

GMATGuruNY wrote:
pnk wrote:Hi Mitch,

I agree with your logic but why my approach gives me a different answer:

Probability of couple never being together = 1 - Probability of couple always together.

Couple AB, CD; Individual E

Arrangement with AB, CD together = 3! 2! 2! (3! bcoz AB, CD, E is 1 group each - and these 3 groups can be arranged among themselves in 3! ways; 2! each for AB and CD arranging among themelves) = 24

So, probability = 1 - 24/120 = 4/5. Could you pls explain where I am at fault.
You need to account for all the different ways that one or both couples could sit together. Here's an equation that you could use:

P(neither couple together) = 1 - P(only AB together) - P(only CD together) - P(both AB and CD together)

Number of ways to arrange AB, CD and E = 3! * 4 = 24.
So P(AB and CD together) = 24/120 = 1/5.

Number of ways to arrange AB, C, D and E with CD not together = (Total arrangements of AB, C, D, and E) - (number of arrangements with both AB and CD together) = (4! * 2) - 24 = 24.
So P(only AB together) = 24/120 = 1/5.

Number of ways to arrange CD, A, B and E with AB not together = (Total arrangements of CD, A, B, and E) - (number of arrangements with both AB and CD together) = (4! * 2) - 24 = 24
So P(only CD together) = 24/120 = 1/5.

Thus P(neither couple together) = 1 - 1/5 - 1/5 - 1/5 = 2/5.

Hope this helps!

Thank u very much Guru, i also done similar mistake as others done after explaining the problem now i can understood how to do....

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neerajbansal: Senior | Next Rank: 100 Posts; Posts: 61; Joined: Fri Feb 19, 2010 10:12 am; Thanked: 3 times

by neerajbansal » Sun Aug 15, 2010 7:20 pm

gmat guru,

What am I missing in my explanation ? Can you explain ?

Q. TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??

--> p(event) --- Desired Outcomes/Total possible

Total Possible --> 5!

Desired Outcomes --> Assume 4 couples are ties with a rope and they are collectively called CCCC. so they can sit with the Single guy (S) in 2! ways. and CCCC can sit together in 4! ways.

So 4!2! is the way 4 couples can sit together with S and they are all together.

So P(4 couples sitting next to each other) is 4!2!/5!

So P(4 couples NOT sitting next to each other) = 1- P(4 couples sitting next to each other)

1 - 2/5

3/5 ans

Or should I think of a couple as a single entity

With that in mind:

c1-c2-s

The number of ways the couples are always next to each other are : 3! * 2!* 2!

so P(that all couple are next to each other) -- > (3! * 2!* 2! )/5! --> 1/5

So P(4 couples NOT sitting next to each other) = 1- P(4 couples sitting next to each other)

1- 1/5

4/5 ans..

CAN YOU TELL ME WHERE AM I GOING WRONG Sad

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