If there's 12 students, why do we go from 11 factorial?

The Blade Dancer · Aug 16, 2020, 2:21 AM

debbie · Aug 16, 2020, 2:24 AM

@The-Blade-Dancer Thank you for asking this question! Prof. Loh uses a multi-step trick here which is very neat! Let's start by reviewing the question:

\text{ We have 12 students} \\ \text{ We have three grades to distribute: A's, B's, and C's} \\ \text{ We need to have at least one of each grade.} \\

Instead of imagining choosing grades to give to each student, we are going to do something a bit weird: we are going to choose students to give to each grade! It's a bit unnatural to do it this way, but this makes it easier to establish the condition that we have at least one of each grade.

Imagine we have three bins: one bin for the A's, one bin for the B's, and one bin for the C's.

We are going to put the people into the bins based on a pre-decided score cutoff (the range of scores for an A, the range of scores for a B, the range of scores for a C). This score cutoff is the curve. We want to count the number of different curves, not the number of ways to give grades to each student. Here I've put scores for each person to show that their order in the line, from "highest score" to "lowest score" is fixed.

Here is an example of one curve:

And here is another example of a different curve:

Notice that the order of the people, arranged by score, is the same regardless of what the curve is. Instead of worrying about the individual grades, we can draw a diagram with just identical people and partitions dividing the different grades of A, B, and C.

Any way to arrange $12$ identical people and $2$ identical partitions corresponds one-to-one with a way to devise a scoring curve.

Oops, but there's another condition we need to satisfy: we need one of at least each grade. So the below diagram wouldn't work:

Cleverly, Prof. Loh showed us that we can pre-emptively assign a person to each of the three "bins." The person with the $100\%$ score will always be in the $A$ bin, and the person with the (sniff, sniff) $57\%$ score will always be in the $C$ bin. We can treat them as identical people figures for the purposes of this calculation.

There are then only $9$ people who are free to be moved from bin to bin, and there are also $2$ partitions, giving us the $11$ objects that we need to rearrange.

Arranging the $11$ objects, $9$ of which are identical people and $2$ of which are identical partitions, has

$\frac{11!}{9! \text{ } 2} = \frac{11 \times 10}{2} = \boxed{55 \text{ ways } }$

This approach is affectionately called "stars and bars ," or "sticks and stones ," among many other names. It's one of the most often used tricks in combinatorics, and I guarantee that you will see it again! It's great to use, as it simplifies problems which otherwise would require a lot of painstaking and error-prone casework!