Timeline for Is there a formula for the number of trees with this extra condition?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2022 at 13:42 | comment | added | Ben Deitmar | ah, yes you're right. I got mixed up. | |
| Apr 12, 2022 at 13:35 | comment | added | Peter Taylor | @Tardis, shouldn't it be $\binom{n-1}{n_1-1}$? | |
| Apr 12, 2022 at 13:24 | comment | added | Ben Deitmar | small correction: we still need to multiply with $\frac{1}{2}{n \choose n_1}$, since for each spanning tree of $K_{n_1,n_2}$ the sets $U,W$ are always $\{1,...,n_1\}$ and $\{n_1+1,...,n\}$ respectively. We want to allow for $U$ to be any subset of $\{1,...,n\}$ with $n_1$ elements (thus the ${n \choose n_1}$ factor) but it still must contain $1$, thus the factor $\frac{1}{2}$. | |
| Apr 12, 2022 at 13:02 | vote | accept | Ben Deitmar | ||
| Apr 12, 2022 at 13:02 | comment | added | Ben Deitmar | Nice, an interesting approach. Thanks! | |
| Apr 12, 2022 at 12:43 | history | undeleted | Racheline | ||
| Apr 12, 2022 at 12:43 | history | edited | Racheline | CC BY-SA 4.0 |
deleted 190 characters in body
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| Apr 12, 2022 at 12:02 | history | deleted | Racheline | via Vote | |
| S Apr 12, 2022 at 12:00 | review | First answers | |||
| Apr 12, 2022 at 12:04 | |||||
| S Apr 12, 2022 at 12:00 | history | answered | Racheline | CC BY-SA 4.0 |