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Ben Brink
Posts:
196
From:
Rosenberg, TX
Registered:
11/11/06


RE: number of onto functions
Posted:
May 19, 2012 5:09 AM



Mahesh, 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. Ben
> 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?



