Sorry my ability of maths is not so good, could anyone could help me to solve this problem, thanks
There are 3 (k=3) countries teams
(eg, china, japan, korea)
Each team has 5 members.
These ppl need to line up in a row.
But each person should have at least one team member beside him.
How many number of ways can they arrange??
How about the general case when k=n countries?
Spent long time on this. Is this from a university course? And if so, what year? I'm curious. This a reasonably difficult problem.
We have:
AAAAA
BBBBB
CCCCC
To obtain at least one member beside another team member they must in:
1) separate pairs and triplets, otherwise at least one will get left out in the cold
2) 5 in a row.
So now we have:
(AAA) (AA)
(BBB)(BB)
(CCC)(CC)
So we now have 6 groups to be arranged
1) 6 choose 2 to place the A's
this also counts the 5 in a row
2) 4 choose 2 to place the B's
3) The C's fall in line. i.e. 2 choose 2
4) 3! to flip the A,B,C's around
this also allows 5 B's in a row and 5C's in a row.
5) Now arrange each of the 5 A's in 5! ways - same for the B's and C's
Finally: (6 choose 2) * (4 choose 2)*(2 choose 2) *(3!) * (5!)^3
First of all, thank for your message.
But I think there is a little wrong in your step 5.
Eg, if AACCC, then ACACC is one of the case among the 5!.
But the second A has no team member
@boyhappy_5 Hmmm, ... let me think about that. My intention is only to permute a group of 5 like elements which may or may not be 5 in a row - not a set set of 5 mixed elements.
_ A1 A2_ _ _ _ _ A3 A4 A5_ _ _ _