Timeline for Solving circular arrangement problem with k identical and m distinct positions.
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 15, 2019 at 3:26 | comment | added | Satish Ramanathan | Your prior argument is not correct. You will have to multiply 10 to 10P9 and then add 10P10 | |
| Nov 13, 2019 at 16:53 | vote | accept | rsonx | ||
| Nov 13, 2019 at 16:53 | comment | added | rsonx | I think you are correct in the first part. There are 10 ways to make someone sit on the distinct chair and 10P9 ways to permute 9 people on 10 identical chairs. Overall ways of arrangement (10.10P9). Anyway, any good book on Counting and Probability containing problems with solutions? | |
| Nov 13, 2019 at 16:28 | comment | added | rsonx | For (iii) I can make someone sit in the distinct chair and permute other 9 people in 10 identical chairs in 10P9 ways OR make no one sit in distinct chair and permute 10 people in 10 identical chairs in 10P10 ways. The presence of an empty distinct chair make the entire arrangement linear, so no need to fix someone on any identical chair. Overall ways of arrangement: 10P9+10P10. Is this correct? | |
| Nov 13, 2019 at 15:48 | history | edited | Satish Ramanathan | CC BY-SA 4.0 | added 22 characters in body |
| Nov 13, 2019 at 15:46 | comment | added | N. F. Taussig | In part I, fix the empty chair. Then the ten people can be arranged in the remaining ten chairs in $10!$ ways. | |
| Nov 13, 2019 at 15:38 | comment | added | awkward | I think I edited my comment while you were responding. It seems to me the way to treat the problem is that there are $10$ people plus one "dummy", for a total of $11$. Then $11$ objects can placed around a circular table in $10!$ ways. | |
| Nov 13, 2019 at 15:35 | comment | added | Satish Ramanathan | Simply one chair is empty, I thought it does not matter that there are 11 chairs for the first part. These chairs are around a circular table and there are 10 people to be arranged in 10 chairs around the table. | |
| Nov 13, 2019 at 15:33 | comment | added | awkward | But the OP says there are $11$ chairs around the table, not $10$. So it seems to me that there are $10!$ possible arrangements (and the answer key is wrong), thinking of the $11$ chairs as seating $10$ people plus one "dummy". | |
| Nov 13, 2019 at 15:29 | history | answered | Satish Ramanathan | CC BY-SA 4.0 |