Timeline for Simple combinatorics problems using division
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 15, 2021 at 13:55 | comment | added | Ilmari Karonen | Another (essentially equivalent) symmetric approach is to note that choosing $k$ out of $n$ distinct items is equivalent to arranging all the $n$ items in a line (in one of the $n!$ possible orders), taking the first $k$ items and forgetting their order (i.e. divide by $k!$), and also forgetting the order of the remaining $n-k$ items (i.e. divide by $(n-k)!$). | |
| Nov 15, 2021 at 13:03 | comment | added | Steven Gubkin | Perhaps A and B are identical twins... | |
| Nov 15, 2021 at 12:42 | comment | added | user21820 | @StevenGubkin: Well there is a simpler form of "1 before 2". How many ways can you put A,B on a row of n seats so that A is on the left of B? Well, n·(n−1)/2. But it's again not really equivalence classes, though the flavour is there. | |
| Nov 15, 2021 at 12:36 | comment | added | Steven Gubkin | Ya, the answers so far seem to advanced. I am wanting very simple examples, like in my original post. | |
| Nov 15, 2021 at 12:15 | comment | added | user21820 | @StevenGubkin: I know, but that seems like the simplest already haha. I cannot imagine trying to get students to understand the number of non-monochromatic necklaces of beads of $p$ colours if they cannot understand $n$ choose $k$. But I guess you might like the $1$ before $2$ example in my other post, since it is just a bit simpler than $n$ choose $k$. =) | |
| Nov 15, 2021 at 11:53 | comment | added | Steven Gubkin | Indeed, the penultimate paragraph in the OP indicates that this is the reason I want simpler examples: to give some experience with thinking about division before developing the formula for combinations. | |
| Nov 15, 2021 at 8:01 | history | answered | user21820 | CC BY-SA 4.0 |