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Topic: number of onto functions
Replies: 4   Last Post: May 20, 2012 7:11 AM

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Ben Brink

Posts: 198
From: Rosenberg, TX
Registered: 11/11/06
RE: number of onto functions
Posted: May 19, 2012 5:09 AM
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On this I'm getting a much bigger answer. Suppose for the sake of argument that each employee may be assigned up to 4 jobs. If the
sequence of numbers of jobs from best employee to worst is (4,1,1,1) then that assignment can be done in 7C4 x 3C1 x 1C1 x 1C1 ways.
Therefore we can assign 4 jobs to 1 employee and 1 job each to the other 3 in 4(7C4 x 3C1 x 1C1 x 1C1) ways.
Similarly, we can assign 3 jobs to 1 employee, 2 jobs to another employee and 1 job each to the other 2 in 12(7C3 x 4C2 x 2C1 x 1C1) ways. Finally we can assign 2 jobs to each of 3 employees and 1 job to the 4th employee in 4(7C2 x 5C2 x 3C2 x 1C1) ways.
Adding the three results above gives me over 7000 possible assignments of jobs. I suggest you try these calculations and decide whether they break down the problem as you understand it.
Thanks and will work on the "word" problem the next couple of days.

> Date: Sat, 19 May 2012 02:43:20 -0400
> From: discussions@mathforum.org
> To: discretemath@mathforum.org
> Subject: number of onto functions
> Q: In how many ways can 7 different jobs be assigned to 4 different employees so that each employee gets atleast one job and the best employee gets the toughest job.
> I suppose this is equivalent to finding number of onto functions from a set of 6 elements to a set of 3 elements, which I get as 540. Am I right?

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