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An entrepreneur wants to assign 5 different tasks to 3 of his employees. If every employee is assigned at least 1 task, how many ways can the entrepreneur assign those tasks to his employees?

My solution is select 3 jobs and assign them to 3 employees. This can happen in $\binom{5}{3}3!$ ways. This ensures that each employee is assigned at least 1 task. After this, the remaining 2 tasks an be assigned to any of the 3 employees. So the final answer becomes $\binom{5}{3}3!3^2$. But the correct answer is 150. So where am I going wrong?

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  • $\begingroup$ As Michael Biro said, you are over-counting. Do you want to know how to get the correct answer though? $\endgroup$ – svelaz Jan 13 '17 at 16:33
  • $\begingroup$ @svelaz Yes please post your solution and it would be nice if you can post a particular case that I am counting more than one. $\endgroup$ – Stupid Man Jan 13 '17 at 16:36
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Either (a) one employee is assigned three tasks and the other two are assigned one task each, or (b) one employee is assigned one task, and the other two are assigned two tasks each.

In case (a) you can choose the special employee in $3$ ways, he may choose $3$ tasks in ${5\choose 3}$ ways, and you can distribute the two remaining tasks to the other two employees in $2$ ways. Gives $3\cdot{5\choose 3}\cdot 2=60$ possibilities.

In case (b) you can choose the special employee in $3$ ways and can choose a task for him in $5$ ways. You can then pair off the remaining four tasks in $3$ ways and distribute the pairs on the the two remaining employees in $2$ ways. Gives $3\cdot 5\cdot 3\cdot 2=90$ possibilities.

The total number of admissible assignments therefore is $150$.

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You are over-counting the ways each employee can be assigned more than one job. For example, if step one assigns task 1 to employee A and then step 2 assigns task 2 to employee A vs the reverse. Both give task 1 and 2 to A but your procedure counts that as 2 different ways.

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  • $\begingroup$ Yes I know stars and bars. If the tasks were identical the answer to this would have been $\binom{4}{2}$ using stars and bars (positive integer solutions of $x+y+z = 5$) $\endgroup$ – Stupid Man Jan 13 '17 at 16:26
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    $\begingroup$ Sorry, stars and bars was the wrong hint to give as the tasks are distinct. Instead, there are only 2 ways to distribute the tasks 2,2,1 or 3,1,1, and calculate from there. $\endgroup$ – Michael Biro Jan 13 '17 at 16:30

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