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 teammember 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_ _ _ _