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2 semi hard QR problems..

Discussion in 'DAT Discussions' started by theedaddy77, Jun 13, 2008.

  1. theedaddy77

    theedaddy77 2+ Year Member

    May 9, 2008

    12 people and you want to sit 6 of them on a round table ?

    12 people and you want to sit 6 them on a bench?

    u should get different answers? can anyone xplain why + show math...
    Last edited: Jun 13, 2008
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  3. doc3232

    doc3232 7+ Year Member

    Feb 15, 2008
    If you sit on a round table, lets say 123456
    It will actually look like 123456123456123456123456 etc
    So in that seating you can "extract" (Chem guy...) 234561 from it and etc
    So thats that.
    I don't know if A am answering correctly :)
  4. vvvv

    vvvv 2+ Year Member

    May 12, 2007
    yes, they should be different. sitting 6 people on the round table is like sitting 5 people on the bench. because the first person sit on the round table, anywhere is doesnt matter but on the bench, its matter.

    for round table. choose 6 from 12 is 12P6= 12!/6! * 6!
    then sit 1 on the table first, you have 5 left. and you have 5! to rearrange 5 people on the table. so the answer is 12P6 + 5!

    for the bench: 12P6 + 6!
    Last edited: Jun 13, 2008
  5. theedaddy77

    theedaddy77 2+ Year Member

    May 9, 2008
    yeah first of all u will do a perm

    for the bench it will have more outcomes than the circle....
    12!/ (12-6)! = 12!/6!

    for the circle
    12!/ (12-5)! = 12!/7!

    basically for a circle imgine that you are choosing one less person. PEACE OUT

    p.s hopefully i got this problem right............ anyone can confirm?
  6. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    Oct 25, 2006
    1. (12 P 6) / 6 = 12!/6!*6 = (12*11*10*9*8*7) / 6 = 110,880
    2. (12 P 6) = 12!/6! = (12*11*10*9*8*7) = 665,280

    For a bench (easier to describe): You want to sit one of the 12 in the first spot, then one of the remaining 11 in the second spot, one of the remaining 10 in the third spot, ..., and one of the remaining 7 in the sixth spot. Order matters here so the quick way is to use a permutation (12 P 6) and you get your answer.

    For a round table you use the same idea as above. Choose an arbitrary starting point and you can put any of the 12 people there. Go around the table and end with the final spot, where you can put any of the remaining 7 people. Again, order matters so a permutation is used (12 P 6). The difference here is that you have a round table (note that it doesn't have to be 'round', it can be square or rectangular or whatever so long as it loops around and back to the beginning). Imagine I choose people A, B, C, D, E, and F and sit them around a table so that person A is at the 12 o'clock position and B through F are in that order clockwise. That's 1 of the 665,280 possibilities. Now imagine I shift everyone one seat clockwise but keep the order. That's 2 of the 665,280 possibilities. By repeating this a few more times, you come to realize that 6 of the 665,280 orders come from seating A-F in one order around the table. Since there are 6 ways to position the first person (6 seats), you only need to consider 1 out of the 6 ways because we consider them to be the same.

    I know their physical positions are different so you might be wondering why this is different from the bench. Well just remember that when people sit on a straight bench, the first and last people are not next to each other. So depending on where you are sitting, you may be next to 1 person or 2 people. At a round table, you are always between two people.

    So since 1 out of 6 are unique, you take 665,280 / 6 and get 110,880.
  7. supraman

    supraman Boston Celtics 7+ Year Member

    Oct 30, 2005
    What did you get in QR?
  8. Streetwolf

    Streetwolf Ultra Senior Member Dentist 7+ Year Member

    Oct 25, 2006
    28 (100 percentile)

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