Sponsored Ad:
Results 1 to 7 of 7

Thread: Difficult Permutation Problem

  1. #1
    Trying to make mom and pop proud
    Join Date
    Apr 2012
    Posts
    6
    Rep Power
    9


    Good post? Yes | No

    Difficult Permutation Problem

    Sponsored Ad:
    I'm having a lot of trouble making my answer look like the OA. Can someone help?

    Here's the problem:

    In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?




    OA to follow after some discussion

  2. #2
    Within my grasp! thepillow's Avatar
    Join Date
    Jul 2012
    Location
    pennsylvania, us
    Posts
    147
    Rep Power
    9


    Good post? Yes | No
    Quote Originally Posted by choked View Post
    I'm having a lot of trouble making my answer look like the OA. Can someone help?

    Here's the problem:

    In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?




    OA to follow after some discussion
    It depends on whether or the each seat at a table is unique. What I mean is that if we consider the tables to have chair1, chair2, chair3, etc... we get a different answer than if we just think of the tables as having 5 or 7 chairs. In other words, does seating the people in the same order but shifting everyone over to the right or the left count as the same arrangement or not?

    If we're only concerned about which people are next to each other and not which particular chairs they occupy, then we have 12c5 (or 12c7) ways to divide them into groups of 7 and 5. Then we have 6! ways to arrange the people at the larger table and 4! ways to arrange the people at the smaller table.

    This gives us (12c5)*6!*4!, which will be a huge number.

    If we care about which chairs they occupy, then the 6! and 4! become 5! and 7!. They would cancel out with the denominator in 12c5, so we would just get 12!.

    These calculations seem right to me for these 2 different cases. What do you think?

  3. #3
    God loves the steadfast Javoni's Avatar
    Join Date
    Aug 2012
    Location
    Tajikistan
    Posts
    13
    Rep Power
    9


    Good post? Yes | No
    There could be an umpteen combination:
    12C5 + 12C7 = calculate it, that is onerous! But what does the problem requires? let 7 people sit across first table and the left 5 people sit across other table? Or otherwise?
    Life begins at the end of your comfort zone

  4. #4
    Trying to make mom and pop proud Nitwick's Avatar
    Join Date
    Jul 2013
    Posts
    2
    Rep Power
    8


    Good post? Yes | No
    Quote Originally Posted by choked View Post
    I'm having a lot of trouble making my answer look like the OA. Can someone help?

    Here's the problem:

    In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?




    OA to follow after some discussion
    You can choose 7 people from 12 in 12C2 ways.
    7 people can be seated in 7! ways
    Remaining 5 can be seated in 5! ways

    Total no of ways = 12C2 * 7! * 6!

  5. #5
    An Urch Guru Pundit Swami Sage
    Join Date
    Aug 2007
    Posts
    696
    Rep Power
    16


    Good post? Yes | No
    Since the question has made no mention of which types of tables these are (ie. a round table will have a different number of distinct seating arrangements than a rectangular one) or the order of how they're seated at the table, I'm assuming that this is just a simple combinations problem. Since every distinct group of 7 people corresponds to an equally distinct group of 5, the answer is simply 12C7 = 12C5 = 12*11*10*9*8/120 = 792

    The only other answer that I would consider (albeit loosely based on the vague nature of the question) would be 2*792 = 1584 (assuming each table can hold either 7 or 5 people)

  6. #6
    Trying to make mom and pop proud
    Join Date
    May 2015
    Posts
    17
    Rep Power
    6


    Good post? Yes | No

    Re: Difficult Permutation Problem

    In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?

    Ans: The first sit can be taken by anyone of 12 people, the next sit by the remaining 11 people and so on. The total number of such arrangements is: 12!
    In my humble opinion, it’s immaterial how we subdivide these arrangements into separate groups be that (5 people at the first table and 7 at the second) or otherwise, or say (6 people at the first table and 6 at the second) still the total number of such arrangements will not change.

  7. #7
    An Urch Guru Pundit Swami Sage
    Join Date
    Aug 2007
    Posts
    696
    Rep Power
    16


    Good post? Yes | No

    Re: Difficult Permutation Problem

    Quote Originally Posted by Tristar View Post
    In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?

    Ans: The first sit can be taken by anyone of 12 people, the next sit by the remaining 11 people and so on. The total number of such arrangements is: 12!
    In my humble opinion, it’s immaterial how we subdivide these arrangements into separate groups be that (5 people at the first table and 7 at the second) or otherwise, or say (6 people at the first table and 6 at the second) still the total number of such arrangements will not change.
    If the order of the individual seat locations matters, then you are correct: the answer is 12!

    Note that this number can also be derived as follows:

    # of Groups of 7 (and 5 left over) that can be created from a population of 12 = 12C7 = 12C5 = 12! / (7! * 5!)

    # of unique arrangements of 7 people in 7 distinct chairs = 7!
    # of unique arrangements of 5 people in 5 distinct chairs = 5!

    # of total unique seating arrangements = 12C5 * 7! * 5! = 12!

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Similar Threads

  1. Another Permutation problem
    By LIV in forum GRE Math
    Replies: 4
    Last Post: 01-27-2009, 10:39 AM
  2. permutation problem
    By e.cartman in forum GRE Math
    Replies: 12
    Last Post: 09-01-2008, 07:52 AM
  3. permutation problem
    By diwakardiwa in forum GMAT Problem Solving
    Replies: 2
    Last Post: 07-27-2008, 05:36 PM
  4. Permutation Problem
    By divi_kk in forum GMAT Problem Solving
    Replies: 1
    Last Post: 01-16-2007, 05:55 PM
  5. Permutation Problem
    By PennFan2002 in forum GMAT Math
    Replies: 1
    Last Post: 02-28-2005, 02:32 AM

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •