# Why is my solution to this counting problem incorrect?

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?

• As Michael Biro said, you are over-counting. Do you want to know how to get the correct answer though? – svelaz Jan 13 '17 at 16:33
• @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. – Stupid Man Jan 13 '17 at 16:36

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 $150$.
• 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 $x+y+z = 5$) – Stupid Man Jan 13 '17 at 16:26