**A woman nearly became violent in a math class I taught several years ago**. I was going over homework problems and she wanted to know whether a certain problem was a “permutation” or a “combination.” She knew how to solve two kinds of problems and was irritated when I told her that her homework problem didn’t fall into either of her two categories.

She insisted that I tell her which of the two techniques would solve the problem and nearly lost control when I repeated that neither would work. The rest of the students and I were shocked. A little nervous laughter broke the tension and we resumed going over the homework.

The angry student had implicitly come to believe that** if a counting problem contains two numbers, n and k, there are only two possible answers**: *P*(*n*, *k*) and *C*(*n*, *k*). Here *P*(*n*, *k*) = *n*!/(*n*–*k*)! and *C*(*n*, *k*) = *P*(*n*, *k*)/*k*!. She was not alone. Students commonly believe this, and for good reason: most homework problems can be solved this way. For example, a club with 12 members can elect five distinct officers in P(12, 5) ways and they can select a committee of five members in *C*(12, 5) ways. It’s easy for an instructor or textbook author to think of dozens of homework problems in these patterns and unintentionally imply that these are these are the only possibilities. However, the following problem contains the numbers 12 and 5 but the solution is neither *P*(12, 5) nor *C*(12, 5).

Suppose you have a class of 12 students. Each student will receive one of five letter grades: A, B, C, D, or F. At the end of the course, you tally up how many students received each grade. How many different ways could the tally turn out? For example, one possibility would be all A’s. Another would be three A’s, four B’s, four C’s, no D’s, and one F.

The grade tally problem is representative of a class of problems that come up fairly often in application but that lie just outside what students typically learn. The general solution is written up in these notes on counting selections with replacement. The notes include the famous “stars and bars” explanation by William Feller.

By the way, there are 1,820 possible grade tallies for 12 students and five grades.

**Related links**:

Is that the problem you gave in class? If so, I hope you at least taught how to solve it. Its not uncommon for teachers to do jack nothing all year long and expect the students to self-teach out of library books. It is your explicit job as a teacher to teach, not just tally grades and file them away with academia, as is oh so common among so-called “prestigious” schools. More like for-profit corporations who sell degrees, if you ask me. If you did teach this subject matter then your student should have known it. Im forced to ask how its possible the student could only assume it was one or the other if she had ever been taught a third or fourth way.

I have a combinatorics question for you whose solution has thus far escaped me.

There are ‘n’ objects. They are not necessarily unique. Unique element number one has a quantity of r₁, unique element number two has a quantity of r₂, so on so forth. In total, r₁ + r₂ + r₃ +… = n

How many different

uniquepermutations exist…. IF I take ‘k’ objects at a time?This was an attempt on my part to integrate problems involving permutation of n objects taken all n at a time with redundant elements, for which the solution is n!/(r₁! r₂! r₃!), and questions involving regular permutations without redundant elements taken k at a time, for which the solution is nPr. I have combined these two concepts into one question and Im stumped as to how to solve.