- #1

NatFex

- 26

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Moved from a technical forum, so homework template missing

## Homework Statement

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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?

## Homework 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: n

^{k}

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

## 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

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