The question is as follows:

A university mathematics department has to assign $10$ professors to $20$ courses. Each professor will only teach $2$ courses, and no courses are repeated. Only one professor will be assigned to teach each course and any professor can teach any course. Find the total number of possible ways that the professors can be assigned to the courses.

$\underline {My \;approach}$:

I tried tackling the problem by assigning a different professor to each of the first $10$ courses.

I then assigned the remaining professors in order for the last $10$, because during the first $10$ courses there are $10$ possibilities for each, I gave that a total of $10^{10}$.

The last $10$ have the number of possible professors reduce each time one is assigned, giving a total of $10!$. So multiplying the numbers we get $10^{10}\times 10!$. Is this reasoning correct?

Thanks for the help!

  • $\begingroup$ Just in case you're wondering about what the problem with you derivation was, there may be professors that only teach courses in the second half. $\endgroup$ – G. Bach Oct 5 '15 at 12:45

I am not able to understand your argument but I don't think there will be a $10^{10}$ term in the answer.

The number of ways you can assign 20 courses to 10 professors is same as the number of combinations of the word AABBCCDDEEFFGGHHIIJJ. Each alphabet is a professor and each place is a course.

So, the answer is $\frac{20!}{(2!)^{10}}$


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