Question 1186714: There are four visitors in a party. There are two tables with two seats each. In how many different ways can these guests sit down?
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website!
Suppose the visitors are A,B,C,D.
4 ways to pick a seat for A
times
3 ways to pick one of the remaining 3 seats for B.
times
2 ways to pick one of the remaining 2 seats for C.
times
the 1 way to pick the only remaining 1 seat for D.
Answer: (4)(3)(2)(1) = 4! = 24 ways.
If the 4 seats are numbered from 1 to 4, here are all 24 ways
they can be seated in the 4 seats.
A B C D
1. 1 2 3 4
2. 1 2 4 3
3. 1 3 2 4
4. 1 3 4 2
5. 1 4 2 3
6. 1 4 3 2
7. 2 1 3 4
8. 2 1 4 3
9. 2 3 1 4
10. 2 3 4 1
11. 2 4 1 3
12. 2 4 3 1
13. 3 1 2 4
14. 3 1 4 2
15. 3 2 1 4
16. 3 2 4 1
17. 3 4 1 2
18. 3 4 2 1
19. 4 1 2 3
20. 4 1 3 2
21. 4 2 1 3
22. 4 2 3 1
23. 4 3 1 2
24. 4 3 2 1
Edwin
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
It depends on what you consider as different ways and what you consider as identical ways.
Had the problem be formulated correctly and in full, it would determine it in the description.
For example, placing (Alice,Bob) and (Bob,Alice) at the same table are different or identical ?
.................
Otherwise, for what reason the tables do present in this problem with two seats each ?
The story should be full, logical and consistent, otherwise it makes strange impression . . .
|
|
|
