Painting a 2x2 Grid We have a 2x2 grid and 10 different colours. I want to paint such that adjacent grids are painted with different colors. How many ways can i do this?
I labeled the grids as:
A B
C D

Scenario 1: B and C different colors.
1)A has 10 colors to choose from.
2)B has 9 colors to choose from
3)C has 8 colors to choose from
4)D has 8 colors to choose from
Total ways: 10*9*8*8 = 5760 ways

Scenario 2: B and C have same colors.
1)A has 10 colors to choose from.
2)B has 9 colors to choose from
3)C has 9 colors to choose from
4)D has 9 colors to choose from
Total ways: 10*9*9*9 = 7290 ways

Total ways = 7290 + 5760 = 13050 ways.

Did i do it correctly?

 A: As Jennifer pointed out in a comment, you have one excess factor $9$ in scenario $2$. Correcting that yields $10\cdot9\cdot8\cdot8+\cdot10\cdot9\cdot9=90(64+9)=6570$.
Another way to do this is via inclusion-exclusion. We have four conditions, one for each pair of adjacent cells. There are $10^{4-k}$ options that fulfil $k$ conditions, except for $k=4$ there are $10$ options instead of $1$, for a total of
$$
9+\sum_{k=0}^4(-1)^k10^{4-k}\binom4k=9+10^4\left(1-\frac1{10}\right)^4=9+9^4=6570.
$$
Another solution: Start at A with $10$ options and go clockwise, with $9$ options per step, which makes $10\cdot9^3$. Then subtract the configurations in which A and C have the same colour, of which there are $10\cdot9\cdot8$, for a total of $10\cdot9(81-8)=6570$.
If you just want a rough estimate, start at A and go clockwise; by the time you get back to A the information which colour you started with will almost be lost, so the probability that you end up with the same colour is almost exactly $1$ in $10$, which effectively reduces the options for the first colour from $10$ to $9$, yielding an estimate of $9^4=6561$.
See also In how many ways can we colour $n$ baskets with $r$ colours?
