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We have 4 different jobs to be done, and a group of 12 people from which to choose workers. How many different ways are there to choose 3 workers for the 4 jobs assuming that one worker does two jobs?

My thinking is that there are ${12}\choose{3}$ ways of choosing the 3 people from the group. Then there are $4\choose2$ ways of selecting two jobs to be merged into one job. Finally there are $3!$ ways to distribute jobs to the chosen people, so the product of these numbers should be the answer.

Is this correct? Is there a different approach for this problem?

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Yes, that’s fine. There are slightly different ways of thinking about it. For instance, you could start by selecting the two jobs to be assigned to one person; as you say, that can be done in $\binom42$ ways. Call the other jobs $J$ and $K$. Now you can pick one of the $12$ for the double job, one of the remaining $11$ for $J$, and one of the remaining $10$ for $K$, for a total of $$\binom42\cdot12\cdot11\cdot10$$ ways of assigning the jobs. Of course

$$\binom{12}33!=12\cdot11\cdot10\;,$$

so this isn’t really all that much different from your approach.

Or you could really stretch it out by first choosing the person to get two jobs, then choosing his pair of jobs, then choosing a person to get one job, then choosing his job, and finally choosing the third worker, who will of course get the only remaining job:

$$12\cdot\binom42\cdot11\cdot2\cdot10\;.$$

But all of these approaches are pretty similar: it’s just a question of the order in which you make the choices.

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  • $\begingroup$ I appreciate your help, thanks. $\endgroup$ – R R Oct 17 '13 at 7:15
  • $\begingroup$ @RR: You’re welcome. $\endgroup$ – Brian M. Scott Oct 17 '13 at 7:17

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