1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Elementary combinatorics problem - why am I wrong?

  1. Jun 25, 2010 #1
    6 unique experienced persons
    2 unique inexperienced persons

    How many 3-person combinations if at most one in the group can be inexperienced?

    6*5*4/3! + 6*5*2/2! is the answer.

    My answer was
    6*5*4/3! + 6*5*2/3!

    Why does the book divide by a 2-permutation and not a 3-permutation for the second term (all combinations consisting of 2 experienced persons and 1 inexperienced person)? My (flawed) rationale is that you have three slots to be filled; we have 6 experienced choices for the first slot, 5 remaining experienced choices for the second slot, and 2 inexperienced choices for the last slot. Since, we don't consider order, we divide by a correction factor of 3 slots permuted.
     
    Last edited: Jun 25, 2010
  2. jcsd
  3. Jun 25, 2010 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you have a group consisting of 2 experienced and one inexperienced, there are 6 choose 2 ways of picking the 2 experienced people, and 2 ways to pick the inexperienced one. Try multiplying those out
     
  4. Jun 25, 2010 #3
    Thanks for your answer, but I'm looking for my flaw visually; why am I under-counting in the second term. Why is my last paragraph explanation wrong?
     
  5. Jun 25, 2010 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your main goal is not to take care of order, you want to take care of double counting. Usually these things are the same, but when dividing out you only want to take care of order within each subset of like objects. So here we have one pair of like objects (the experienced people) whose order is irrelevant, meaning you only divide by 2!

    So if the inexperienced people are named Adam and Abe, and some experienced people are Bob, Billy and Bubba:

    One possibility is to pick Bob, Bubba and Adam

    One possibility is to pick Bubba, Bob and Adam. (so the group of Bob, Bubba and Adam has been counted twice)

    Picking Adam, then Bubba then Bob is not a possibilty because you said you were picking two experienced people, then one inexperienced person, so you never counted this possibility to begin with. (so the group was only counted twice, not more times)
     
  6. Jun 25, 2010 #5
    That finally cleared it up for me. Thanks a lot.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook