andre gower

901541688AITOP1 1688AI1537 644 TOP1!
199916881688 1688CBBS.
1688.com
FAQ 11688 21688.
1688 B2B.
1688__ID.
BBBB 1688.com.
1688PC 116881688307.
1688~ 2021/02/12~ 1688
1688 ERP1688 1688.

901541688AITOP1 1688AI1537 644 TOP1!
199916881688 1688CBBS.
1688.com
FAQ 11688 21688.
1688 B2B.
1688__ID.
BBBB 1688.com.
1688PC 116881688307.
1688~ 2021/02/12~ 1688
1688 ERP1688 1688.
Each corresponds to exactly one ordered triple with $ a < b < c $ satisfying the condition.
Question: A topologist labels 6 distinct regions of a cell membrane using 3 red and 3 blue topological markers, assigning one marker per region uniformly at random. What is the probability that no two adjacent regions (in a fixed linear chain) are both colored red?
This is a classic non-adjacent independent set problem.
Let $ f(n,k) $ be the number of ways to place $ k $ non-adjacent red markers in $ n $ positions in a line.
We need $ f(6,3) $ with no two reds adjacent.
Check manually: list all binary strings of length 6 with exactly three 1s (red), no two adjacent.
Such configurations must have at least one blue between reds. So min spacing: R-B-R-B-R, requires 5 positions. We have 6, so one extra red to place.
Start with the base: R-B-R-B-R, and have one extra R to place in the 4 possible gaps: before first, between, or after last.
Actually, the standard transformation: place 3 reds with at least one blue between, using stars and bars with spacing.
Thus, number of favorable outcomes: 4