Timeline for Random Bridge Hand w Cards of exactly two suits
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2023 at 4:25 | comment | added | ensbana | By the same logic, I guess the number of hands with exactly 3 suits would be ${4 \choose 3}{39 \choose 13} - 2({4 \choose 2}{26 \choose 13} - 4 \cdot 3) - 4 \cdot 3$? | |
| Oct 7, 2018 at 17:46 | vote | accept | elcharlosmaster | ||
| Oct 7, 2018 at 17:42 | comment | added | Zubin Mukerjee | Good point, thanks | |
| Oct 7, 2018 at 17:41 | comment | added | TonyK | +1 for this. But slightly simpler is to say that for each choice of two suits, there are $\binom{26}{13}-2$ ways to choose the cards. This gives you the same answer. | |
| Oct 7, 2018 at 17:39 | history | answered | Zubin Mukerjee | CC BY-SA 4.0 |