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willbeatthegmat
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ashish1354
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awesomeusername
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Please see this link so a little more insight on how to approach these types of problems.
https://www.math.utah.edu/~levin/M5010/HW/hw2sol.pdf
https://www.math.utah.edu/~levin/M5010/HW/hw2sol.pdf
Last edited by awesomeusername on Mon Jan 26, 2009 11:26 pm, edited 1 time in total.
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Brent@GMATPrepNow
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Here's how I would set this up:willbeatthegmat wrote:In how many ways can 5 girls & 3 boys be seated in a row such that no 2 boys are together?
Have 11 chairs and first seat the girls in seats 2, 4, 6, 8, and 10
_G_G_G_G_G_
This can be accomplished 5! ways.
Note: This arrangement prevents the boys from sitting together.
Now seat each of the 3 boys in one of the 6 remaining seats (then remove the 3 empty seats).
This can be accomplished in 6x5x4 ways
So, the total number of ways to seat all of the boys and girls is 5! x 6x5x4
Brent Hanneson - Creator of GMATPrepNow.com
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awesomeusername
- Master | Next Rank: 500 Posts
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I have my doubts about the approach mentioned above.
The question states that no two boys should be seated together.
Here's how I approached it. I may be faulty as well.
In these questions its probably better to work backwards, i.e find the number of +ve cases and then subtract from the total to find the number of -ve cases.
Case I
Treat 3 boys sitting together as 1 seat
So the total ways are,
6 * 5 * 4 * 3 * 2 * 1 = 6! ways
Now the boys themselves can be seated in 3! ways within themselves.
Therefore, there are 6! * 3! ways
Case II
Now, we need to find 2 boys sitting together (which is what originally is asked)
Here again treat the 2 boys as 1 seat,
7 * 6 * 5 * 4 * 3 * 2 * 1 = 7! ways
But again there are 2 ways the chosen boys can arrange themselves and another 3 ways the 2 boys can be chosen
So the total number of ways = 7! * 2 * 3 = 7! * 6
[we need to subtract the 12 ways in which all 3 would sit together, which is already included above in Case I]
The total number of ways = (7! * 6) -12 ways
So the total +ve cases are, 6 * 6! + (6 * 7!) -12 = (48 * 6!) - 12
The total possible cases are 8!
So the -ve cases must be = 8! - (48 * 6!) + 12 = (8 * 6!) + 12 ways
Therefore 5772 ways.
The question states that no two boys should be seated together.
Here's how I approached it. I may be faulty as well.
In these questions its probably better to work backwards, i.e find the number of +ve cases and then subtract from the total to find the number of -ve cases.
Case I
Treat 3 boys sitting together as 1 seat
So the total ways are,
6 * 5 * 4 * 3 * 2 * 1 = 6! ways
Now the boys themselves can be seated in 3! ways within themselves.
Therefore, there are 6! * 3! ways
Case II
Now, we need to find 2 boys sitting together (which is what originally is asked)
Here again treat the 2 boys as 1 seat,
7 * 6 * 5 * 4 * 3 * 2 * 1 = 7! ways
But again there are 2 ways the chosen boys can arrange themselves and another 3 ways the 2 boys can be chosen
So the total number of ways = 7! * 2 * 3 = 7! * 6
[we need to subtract the 12 ways in which all 3 would sit together, which is already included above in Case I]
The total number of ways = (7! * 6) -12 ways
So the total +ve cases are, 6 * 6! + (6 * 7!) -12 = (48 * 6!) - 12
The total possible cases are 8!
So the -ve cases must be = 8! - (48 * 6!) + 12 = (8 * 6!) + 12 ways
Therefore 5772 ways.
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Brent@GMATPrepNow
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This approach will work; however, there are a few problems with your calculations:Case II
Now, we need to find 2 boys sitting together (which is what originally is asked)
Here again treat the 2 boys as 1 seat,
7 * 6 * 5 * 4 * 3 * 2 * 1 = 7! ways
But again there are 2 ways the chosen boys can arrange themselves and another 3 ways the 2 boys can be chosen
So the total number of ways = 7! * 2 * 3 = 7! * 6
[we need to subtract the 12 ways in which all 3 would sit together, which is already included above in Case I]
The total number of ways = (7! * 6) -12 ways
In case II you counted each BBB situation twice.
To see how this happened, we'll let the 3 boys be A, B, and C. If we follow your solution and consider the case where A and B are combined into one child (AB), we could have one possible seating arrangement of GGG(AB)CGG. In another situation (which you have counted separately) we can combine B and C together to form one child (BC). In this situation, we can have the seating arrangement GGGA(BC)GG
As you can see, we have counted the arrangement GGGABCGG twice. So, from your total of 7!*6, we need to subtract all of the BBB arrangements since these situations where all 3 boys are seated together have been counted twice. How many BBB situations are there? There are 6!*3! (as calculated in case I)
So, the final value for case II should have been 7!*6 - 6!*3!
Finally, we need to recognize that all of the exceptions (at least 2 boys together) are already accounted for in case II. We don't need to include case I (although we did need it to determine the total value for case II)
So, the final answer is 8! - (7!*6 - 6!*3!)
You will find that this answer is the same as my answer of 5! x 6x5x4
Brent Hanneson - Creator of GMATPrepNow.com
Thanks Brent.
I actually did realise my mistake as I read it through.
But I agree, your method was faster arriving at. Thanks for helping get to understand the methodical way at arriving at your answer.
I actually did realise my mistake as I read it through.
But I agree, your method was faster arriving at. Thanks for helping get to understand the methodical way at arriving at your answer.
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Johnniewales
- Newbie | Next Rank: 10 Posts
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Hi Brent,
Please why are there always 11 Seats when you solve this type
of problem, I noticed you also used 11 seats in the case of 4 Ladies and 5 gentlemen
Seated in a row and no two ladies sit together. please what's the
Secret behind the 11 spaces.
Please why are there always 11 Seats when you solve this type
of problem, I noticed you also used 11 seats in the case of 4 Ladies and 5 gentlemen
Seated in a row and no two ladies sit together. please what's the
Secret behind the 11 spaces.
