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Posers and Puzzles

Posers and Puzzles

  1. 11 Jun '08 22:56
    5 teams with 2 cars each will compete in a race.
    The cars are arranged in two columns in such a way each car in the right column has a car of a different team in the same row of the left column.
    How many different configurations are possible?
  2. 12 Jun '08 02:37
    Hmm...The stipulation does allow to have some cars on the left that don't have a car to its right, correct?
  3. 12 Jun '08 02:45
    No.
  4. Subscriber AThousandYoung
    Proud Boys Beware
    12 Jun '08 02:49
    Originally posted by smaia
    No.
    Yes, it does. You should change it to make this clear. For example, stipulate 5 and only 5 rows.
  5. 12 Jun '08 02:54 / 1 edit
    Originally posted by AThousandYoung
    Yes, it does. You should change it to make this clear. For example, stipulate 5 and only 5 rows.
    Let's avoid confusion.
    the configuration must be:

    XX
    XX
    XX
    XX
    XX
  6. 12 Jun '08 11:53
    i'm not that great at combinatorics, in fact i'm pretty bad, Dejection probably can give the right answer if i'm wrong but my working out is 3219833 (quite big so i think i might be wrong)

    explanation:

    the first position (call it 1,1) can have a total of 10 different cars.
    (1,2) can have a total of 8 different cars. (10-1 -1 that is the same)
    (2,1) can have 8 total cars (10-2)
    (2,2) can have 7 total cars (10-3)
    etc etc
    until (10,2) can have 1 possibly car
    this gives an answer of 3225600

    this isn't quite finished though as this also leaves the possibility that 2,1 and 2,2 etc etc are on the same team

    so you minus
    7 x 6! (8 possible cars that (2,1) can be and 6! is the number of arrangements that follow those cars being on the same team)
    5 x 4!
    3 x 2!
    and 1
    leaving you with an answer of 3219833

    oh and Dejection, if you see this answer is horribly wrong, don't laugh, i don't really like combinatorics or maths so i don't look into any of this any further than i get taught
    also i worked this out on paper so if theres any basic errors then meh
  7. Subscriber deriver69
    Keeps
    12 Jun '08 12:08 / 2 edits
    Are arrangements considered different if the cars in the same team are swapped?

    i.e. is A b /a B/c D/e C/d E differrent to a B/A b/C d/E c/D e ?

    (If it is not obvious the teams are denoted by letters with each car distinguished between higher and lower case)
  8. 12 Jun '08 12:12
    Originally posted by deriver69
    Are arrangements considered different if the cars in the same team are swapped?

    i.e. is A b /a B/c D/e C/d E differrent to a B/A b/C d/E c/D e ?

    (If it is not obvious the teams are denoted by letters with each car distinguished between higher and lower case)
    i assumed so and i'm sure i added it into my proof
    after all, different cars are different cars
  9. 12 Jun '08 14:46
    Originally posted by deriver69
    Are arrangements considered different if the cars in the same team are swapped?

    i.e. is A b /a B/c D/e C/d E differrent to a B/A b/C d/E c/D e ?

    (If it is not obvious the teams are denoted by letters with each car distinguished between higher and lower case)
    Yes.
  10. 13 Jun '08 14:19
    Originally posted by banx99
    i'm not that great at combinatorics, in fact i'm pretty bad, Dejection probably can give the right answer if i'm wrong but my working out is 3219833 (quite big so i think i might be wrong)

    explanation:

    the first position (call it 1,1) can have a total of 10 different cars.
    (1,2) can have a total of 8 different cars. (10-1 -1 that is the same)
    (2,1) ca ...[text shortened]... than i get taught
    also i worked this out on paper so if theres any basic errors then meh
    I got a slightly different variation with the highest possible solution being 3,225,600...

    10 8
    8 7 or 6
    6 5 or 4
    4 3 or 2
    2 1

    ok a little explanation now.. there is obviously 10 cars that can go in the first position that leaves 8 others in the second position (crossing out the first car and its partner), after the first two cars are chosen there remains 8 cars. if this is the partner of one of the above cars then there are 7 cars left to chose from but if it is from a different car duo then there is only 6 choices, this leaves 6 cars for the next row... and so on... then I just multiplied them all together... but what I don't know is how to get rid of the duels in the second column... any ideas? or am I just totally off on my thinking on this problem?
  11. 13 Jun '08 19:15 / 1 edit
    Originally posted by smaia
    5 teams with 2 cars each will compete in a race.
    The cars are arranged in two columns in such a way each car in the right column has a car of a different team in the same row of the left column.
    How many different configurations are possible?
    OK, working my way through this, assuming each car is distinct. Perhaps next post I'll treat both team cars as indistinguishable.

    For the purpose of analysis, I number each pair of cars (1-5), and which car on that team it is (1 or 2). They are numbered before pairing off.

    I then map how the cars are "linked". What I mean by that is this. Car 1-1 is paired with another car. I then look to see who his partner is paired with, and who that person is paired with, and so on and so forth.

    Now there are two possibilities here. Either I get a chain of all 5 teams, or I have 3 teams linked, and the other two are paired with each other's partners.

    An example of the first. would be
    1-1 with 2-2, 2-1 with 3-2, 3-1 with 4-1, 4-2 with 5-2, 5-1 with 1-2

    An example of the second..
    1-1 with 2-1, 2-2 with 3-1, 3-2 with 1-2, 4-1 with 5-1, 4-2 with 5-2

    So now I look at how many ways you can pair off the cars.

    Chain of 5
    I count 24*16 = 384 ways of pairing cars in this manner. (4! * 2^4)

    Chain of 3 + Swapped Partners
    There are 10 combination of 2 teams who can be the ones to "swap", and 2 ways they can swap partners.

    For the remaining 3, we-ll start with the lowest-numbered pair, and examine it as with the chain of 5, which gives 2! * 2^2 or 8 pairings

    Multiply these together for 20*8 or 160 potential pairings of the 5 vehicles.

    Combine the two cases and you have 384+160 or 544 possible sets of pairings which meet the qualifications.

    Front to Rear and Right to Left
    Once all cars are paired off, the pairs are randomly picked to take a row at the starting line, with one of the two picked to be the left car.

    This results in 5! * 2^5 potential arrangements for each set of pairings. 3840 ways to arrange the cars

    This means there are 544 * 3,840 possible arrangements of 5 teams of 2 into 5 rows of two cars where no car is to the side of his racing partner.

    If my math is correct, this equals 2,088,960 distinctly different possible arrangements.
  12. 13 Jun '08 20:00
    My numbers for this problem, but without distinguishing the two cars from a team from each other is this.

    44 pairings (24 chains of 5, 20 chains of 3 + swapped partners)
    3840 arrangements of pairs.

    168,960 configurations.
  13. 13 Jun '08 21:37
    Originally posted by geepamoogle
    OK, working my way through this, assuming each car is distinct. Perhaps next post I'll treat both team cars as indistinguishable.

    For the purpose of analysis, I number each pair of cars (1-5), and which car on that team it is (1 or 2). They are numbered before pairing off.

    I then map how the cars are "linked". What I mean by that is this. Car 1 ...[text shortened]...
    If my math is correct, this equals 2,088,960 distinctly different possible arrangements.
    CONGRATULATIONS!!!
    You got it right!
  14. 14 Jun '08 00:28
    Wow, chains. Never though of that.
  15. 14 Jun '08 13:58 / 1 edit
    dont understand chains, explain
    EDIT: if its overly complicated or pretty simple then dont bother. if its easy enuff ill figure it out. if its too hard, i cbf figuring it out XD