The following 8 sprinters will participate in a fictitious 100 m sprint:

Usain Bolt: Jamaica,

Yohan Blake:Jamaica,

Asafa Powell :Jamaica,

Tyson Gray :USA,

Richard Thompson :Trinidad and Tobago,

Donovan Bailey: Canada,

Olusoji Fasuba: Nigeria, and

Frankie Fredericks: Namibia

If the TWO rules below need to be followed, how many different lane assignments are possible?

- NO two sprinters from the same country can be in adjacent lanes.
- The outside lanes ( 1 and 8) can't be occupied by sprinters from the same country.

The answer is 11520.

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There are 2 cases we have to consider: one of the outside lane is occupied by a Jamaica sprinter and all Jamaica sprinters are on the inside lanes.

Case 1: one of the outside lane is occupied by a Jamaica sprinterFrom the 3 Jamaica sprinters, we choose 1 to stand on a outside lane (Let's say, lane 1) (the number of cases for lane 8 is the same, so just times 2 finally).

Then the other 2 Jamaica sprinters can only stand on lane 3-7 due to the rules, and they cannot stand adjacent to each others, which yield 6 possible arrangements for them (remember, each arrangement yields $2!$ combinations!) $(3, 5), (3, 6), (3, 7), (4, 6), (4, 7), (5, 7)$

For the rest of 5 sprinter, just arrange them randomly on the 5 remaining lanes. Therefore there are $5!$ for these 5 sprinter.

So the possible assignments in this case is $3 \times 2 \times 6 \times 2! \times 5! = 8640$

Case 2: all Jamaica sprinters are on the inside lanesAll Jamaica sprinters can only stand on lane 2-7 without "adjacenting" to each other, which only yields 4 possible arrangements for them (remember, each arrangement yields $3!$ combinations!) $(2, 4, 6), (2, 4, 7), (2, 5, 7), (3, 5, 7)$

Again, the remaining 5 sprinters yield $5!$ combinations on the 5 remaining lanes.

So the possible assignments in this case is $4 \times 3! \times 5! = 2880$

At last, add up all possible assignments, we have $8640+2880=\boxed{11520}$