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!

Permutations (with repetitions) problem

  1. Oct 26, 2016 #1
    • Moved from a technical forum, so homework template missing
    1. The problem statement, all variables and given/known data

    The question is phrased in the following way:

    There are 6 jobs to be assigned to 5 people. Each job is assigned only to one person, and each person must have at least one job. How many different arrangements are there?

    2. Relevant equations

    In general, I would approach a selection of k objects from a total of n in the following way:

    Permutations without repetition: n!/(n-k)!
    Permutations with repetition: nk
    Combinations without repetition: n!/((n-k)!k!) – I express this as nCr(n,k) below.

    3. The attempt at a solution

    Given those conditions, then it stands to reason that 4 people will have 1 job and someone will have 2.

    I approached the problem by thinking about the number of possible employees for every job then multiplying it all out:

    Job 1: 5 (all of them)
    Job 2: 4
    Job 3: 3
    Job 4: 2
    Job 5: 1
    Job 6: 5 (allocated to someone who also did another job above)

    5*4*3*2*1*5 = 5! * 5 = 600

    This seemed to me like it should logically work. The job numbering is obviously arbitrary so this should take into account any assortment/ordering

    However, the correct answer turned out to be 4! * nCr(6,2) = 360. The reasoning behind this revolves around splitting the problem into the first 4 jobs (hence the 4!), then, for the last 2 jobs, working out how many ways there are to select 2 jobs out of 6 for 1 person to do (hence 6 choose 2.)

    This makes a little bit of sense, but not as much as my initially proposed solution. Can someone explain exactly where this succeeds and where mine falls short? Why is my number larger? What are the 'excess' possibilities it's counting? Thanks
    Last edited: Oct 26, 2016
  2. jcsd
  3. Oct 26, 2016 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The problem with your counting method is that jobs 1 & 2 could be done by the same person.

    That said, I don't agree with the "correct" answer.
  4. Oct 26, 2016 #3
    Does it really matter if I arranged it as 5*4*3*2*1*5 or 5*5*4*3*2*1? (In the case of the latter, jobs 1 & 2 can be done by the same person, but given that multiplication is commutative and that, as I've said, the job numbering is arbitrary, does it make a difference?)

    I would like to see your take on the correct answer, if you don't mind.

  5. Oct 26, 2016 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I would do it for 3 jobs and 2 people. It's easy to count them all. Then you can check whether your formula works.
  6. Oct 26, 2016 #5
    Indeed it doesn't! Formula yields 4 whereas there are 6 countable solutions. However, all I'm alerted to is the fact that I'm wrong and not how/why. I realise this is much to ask, so any input is appreciated.
  7. Oct 26, 2016 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You must label the jobs and the people before you start. Then, you exclude the option of jobs 1 & 2 being done by the same person.

    To help you, my starting point was to think that one person must do 2 jobs and the other 4 people must do 1 job each.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Permutations repetitions problem Date
Integration problem using u substitution Monday at 1:02 PM
[Discrete 2] Permutation/Combination Question Feb 13, 2018
Multiplying permutation cycles Sep 1, 2017
Abelian Permutation Group May 26, 2017
Permutations and Transpositions Mar 13, 2017