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Ten nominees for three business awards are randomly seated

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Re: PS: Probability [#permalink]

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New post 15 Aug 2008, 11:12
I'll try. I'll be honest, it didn't make 100% sense to me when I read it in your quote of me!

Remeber that at a round table, #10 and #1 are seated next to each other.

So starting at #1 and taking one from each side gives us 3 together, (remember we're finding out how many DO NOT work so we can subtract that fraction from 1 (or subtract that number from all possible and create the fraction afterwards).

How many ways will Seat # 1 be involved when all 3 seats are in a row?

9-10-1
10-1-2
1-2-3

Now if we consider the same thing for seat #2, we have

10-1-2
1-2-3
2-3-4

Notice, that two of these are idential to what we counted with regard to #1. So, for the first seat, there are 3 options, and for the others, there is only 1 instance that will not create a duplicate. There would be 3 options for every seat as you can see, but the duplicates make it where we can't count 2 of those 3 when considering all seats but the first seat selected to count. For seat #2, 2-3-4 is the only non-duplicate. Counting around the table,

We have

10-1-2
1-2-3
2-3-4
3-4-5
4-5-6
5-6-7
6-7-8
8-9-10
9-10-1

Does this clear it up ?

tarek99 wrote:
jallenmorris wrote:
Wouldn't there be 120 total options? 3C10

\(\frac{10!}{3!7!} = 10*9*8 / 3*2 = 10*3*4 = 120\) and then we must remove out any possible scenarios with seats together.

For each seat, there will be 2 possible ways it can have an adjacent empty set, or 2C3 = 3 ways per seat. But 3 is only for the first one. Lets say you start with Seat #1 at a round table. On either side of Seat #1 is Seat #10 and Seat #2. 2C3 filles the 2 empty seats scenario and is what we must count. But once you have this, if you move to Seat #2, you cannot count #1 next to it as that situation has already been counted when analyzing Seat #1. So, to account for this, when you use #1, you must move to seat # 3 or you risk double counting.

So Seat #1 = 3; #3 = 3; #5=3; #7=3; #9 = 3...that's 15 ways that 2 adjacent seats can be next to each other. Now figure out when all three seats will be next to each other.

This would be (3C5 or 3P5 because order matters?) because the order for Seat #1 could be
I find it less confusing to just list out the number of ways all 3 will be together.

1-2-3
2-3-4
3-4-5
4-5-6
5-6-7
6-7-8
7-8-9
8-9-10
9-10-1
10-1-2

Now about 2 seats together - where x could be any seat except for the seats surrounding the 2 chosen. We need to consider 1-2-4 different than 1-2-5 so we have 6 seats left that can be selected and will not satisfy the 3 seats in a row we've already counted.
1-2-x (*6) [running total]
2-3-x (*6) [12]
3-4-x (*6) [18]
4-5-x (*6) [24]
5-6-x (*6) [30]
6-7-x (*6) [36]
7-8-x (*6) [42]
8-9-x (*6) [48]
9-10-x (*6) [54]
10-1-x (*6) [60]

so we have 70/120...for 7/12.

HA! We may be wrong, but at least we're not alone!

EDIT: Did a google search for the question and found: http://www.manhattanreview.com/forum/showthread.php?p=1217



thanks buddy for your input here, but i'm having a hard time understanding how we can have 3 different possibilities for each said. This is what you said "For each seat, there will be 2 possible ways it can have an adjacent empty set, or 2C3 = 3 ways per seat. But 3 is only for the first one."
would you please explain?
thanks

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Re: PS: Probability [#permalink]

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New post 15 Aug 2008, 11:12
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obviously we cant have MS excel in exam ... and listing all possible configuration is also not possible

here is how it can be done mathematically

total number of ways = 9!

for fav ways lets take out 3 from 10 people. people left = 7
number of ways there 7 people can sit = 6!

now we need to fit these 3 people
first can go at any of the 7 places availabe = 7 ways
second can go at 6 places availabe = 6 ways
third can go at 5 places available = 5 ways

probability = 6! * 7 * 6 * 5 / 9! = 30/72 = 5/12 ... answer

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Re: PS: Probability [#permalink]

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New post 15 Aug 2008, 11:34
well in my opinion excell is not at all required.

If you go by approach : 1-(unfavorable probability)

just take out the no. ways its possible for the adjacent seats to be empty :

2 cases :

1 when all three are adjacent : 10 cases

2 when 2 are adjacent then again 10 case , this multiplied by the no. of possibilities for the third seat i.e 6

=> (10+10*6)/10C3 [10C3 = total no. of events for choosing 3 people out of 10]

= 7/12 [probability for unfavorable events ]

ans = 1-7/12 = 5/12

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Re: PS: Probability [#permalink]

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New post 15 Aug 2008, 11:58
No, excell is generally not required, but after you try to solve it 3 times and come up with 3 different answers, it's best to lay it out nice and easily.

stallone wrote:
well in my opinion excell is not at all required.

If you go by approach : 1-(unfavorable probability)

just take out the no. ways its possible for the adjacent seats to be empty :

2 cases :

1 when all three are adjacent : 10 cases

2 when 2 are adjacent then again 10 case , this multiplied by the no. of possibilities for the third seat i.e 6

=> (10+10*6)/10C3 [10C3 = total no. of events for choosing 3 people out of 10]

= 7/12 [probability for unfavorable events ]

ans = 1-7/12 = 5/12

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Re: PS: Probability [#permalink]

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New post 15 Aug 2008, 14:12
PatrickBateman wrote:
so....uhhh....what is the way to do this <2 min?

Don't think they'll like us busting out Excel on our mobiles at the test center. :P



Take a guess and move on to the next question... most likely it will be an experimental question :|

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Re: PS: Probability [#permalink]

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New post 15 Aug 2008, 14:51
bhushangiri wrote:
PatrickBateman wrote:
so....uhhh....what is the way to do this <2 min?

Don't think they'll like us busting out Excel on our mobiles at the test center. :P



:| Take a guess and move on to the next question... i am not even joking.. questions like this will drain your time. it you dont get it in 3 mins, what is the likelihood that you are going to get it in 2 more ? Most likely it will be an experimental question.

I got this after about half an hour of sweating..
As for actual solution.. total arrangements of 10 in circular permutation = 9!


7 non winners arranged in 6! ways. For the 3 to be not sitting together.. first winner has 7 places to get into, second has 6 of the remaining and third has 5 of the remaining.

6! * 7 * 6 * 5/ 9! = 5/12
*****************************************************************
The second approach, which i tried first and spent 25 of the 30 mins, was obviously the worse of the two. But it paved the way for the better approach.

Total arrangements = 9! as above.

Now for finding the number of ways of seating, when at least two of the 3 winners are together...

Case 1.. all 3 winners are together. 3 winners can be arranged in 3! = 6 ways. This group of 3 can be place in any of the 7 slots between the 7 ppl. So number of ways = 6*7

Case 2.. only two of the 3 are together. Two of the 3 can be arranged in 3P2 = 6 ways. Now we have two groups to place (the duo and the loner). Regardless of which you place first, there are 7 places for the first, 6 for the second. So total ways this can be done = 6*7*6

Total unfavorable cases = 6*7 + 6*7*6
Desired prob = 1 - (6*7 + 6*7*6)/9! = 1-7/12 = 5/12 whew....

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Re: PS: Probability [#permalink]

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New post 15 Aug 2008, 15:54
this is the approach I started with...but I got stuck because I thought that when I said all combinations where 2 are seated together would double count the scenarios where the 3 were sitting together.

stallone wrote:
well in my opinion excell is not at all required.

If you go by approach : 1-(unfavorable probability)

just take out the no. ways its possible for the adjacent seats to be empty :

2 cases :

1 when all three are adjacent : 10 cases

2 when 2 are adjacent then again 10 case , this multiplied by the no. of possibilities for the third seat i.e 6

=> (10+10*6)/10C3 [10C3 = total no. of events for choosing 3 people out of 10]

= 7/12 [probability for unfavorable events ]

ans = 1-7/12 = 5/12

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Re: PS: Probability [#permalink]

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New post 15 Aug 2008, 15:56
bhushangiri wrote:
PatrickBateman wrote:
so....uhhh....what is the way to do this <2 min?

Don't think they'll like us busting out Excel on our mobiles at the test center. :P



Take a guess and move on to the next question... most likely it will be an experimental question :|


I agree. That's why i asked for <2min. If I can't solve it after 2 minutes, I pretty much give up on the question. I spend up to 30 more seconds forming an educated guess.

that being said, I can see a lot of people burning 5 minutes on a strange looking experimental question as you suggested.
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Re: PS: Probability [#permalink]

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New post 18 Aug 2008, 05:47
I can't understand how to approach this kind of questions.

Do we have a method to do this problem in less than 2 minutes??

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Re: PS: Probability [#permalink]

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New post 19 Aug 2008, 08:16
This does not seem to be a typical "circular arrangement" question. The author just wants us to find out the number of arrangements where empty seats are not adjacent to each other.
The 3 empty seats are to be considered "identical" (Regardless of which of the 3 nominees sat on them prior to their getting empty). This is an important fact IMO. Now, out of 10 seats, 3 empty seats can be chosen in 10 C 3 ways, i.e., there is a total of 10 C 3 arrangements of 3 empty seats among 10 total seats. This part is easy to understand, since it is no different than a linear arrangement.
If you arrange adjacent numbers on a circle in pairs, you will come up with 10 pairs. This implies, a pair of empty seats could be arranged in 10 ways around a Round table. Now we are left with 1 empty seat to account for, and this could be 1 among any of the left 8 seats. Therefore, the total arrangements of a pair of empty seats are: 10 C 1 * 8 C 1.

Now, it gets little tricky. E 1, E 2, E 3 are the 3 empty seats and they are identical. All 3 empty seats could also appear adjacent to each other and they could be viewed as a pair of empty seats and an adjacent empty seat e.g., E 1 & E 2 could be a pair and E 3 could be the adjacent empty seat or E 2 & E 3 could be a pair and E 1 could be the adjacent empty seat. This implies that for each triplet there are 2 identical combinations ( Triplet means, 3 empty seats adjacent to each other). Therefore 1/2 of these should be subtracted from the total or else, they will be counted twice in the final tally. So, we have to find out how many such triplets are there.
Again, visualize a circle with 10 numbers. If you arrange them in triplets, you will find there are 10 triplets. Hence there will be 10 * 2 = 20 duplicate combinations. Therefore, we will deduct 10 from the total. The final Mathematical equation should look like this:
10 C 3 - {(10 C 1 * 8 C 1) - 10} / 10 C 3 = 5/12.

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Re: PS: Probability   [#permalink] 19 Aug 2008, 08:16

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