99 to beat

992k120hz 1080P/4k 60hz .
2025
[2026-03-16] [2026-03-25] [2026-03-31] [2026-03-25] .
992418 23.
X 3317870 99 > > v
10# 99717***024073211 2499ms,79.95888ms 588.
MMO
13
gm.99.com - .
[2025-09-30] [2025-08-21] [2025-06-09] 12 [2024-12-12] .

992k120hz 1080P/4k 60hz .
2025
[2026-03-16] [2026-03-25] [2026-03-31] [2026-03-25] .
992418 23.
X 3317870 99 > > v
10# 99717***024073211 2499ms,79.95888ms 588.
MMO
13
gm.99.com - .
[2025-09-30] [2025-08-21] [2025-06-09] 12 [2024-12-12] .
Step 1: Fix the relative order of R's and L's.
The condition is that both R's come before both L's — that is, in the sequence, every R appears before every L. This is more restrictive than just "the two R’s together" or "R’s clustered."
The positions of the 4 R and L flowers (2 R, 2 L) must be such that all R’s occupy the earliest 2 of these 4 positions, and the L’s occupy the last 2.
But since the types are indistinguishable, we count the number of ways to assign positions to the types under the constraint.
Alternative approach:
First, choose 5 positions out of 5 for placing the flowers — but since flower types have repetition, we count the number of distinct permutations where both R’s come before both L’s.
Let’s think combinatorially: Among all distinct sequences of 2 R’s, 2 L’s, and 1 T, how many have all R’s before all L’s?
Let’s denote such sequences as valid if, in order, every R appears before every L.
We can fix the positions of the 2 R’s, 2 L’s, and 1 T, then impose the constraint that the largest R index < smallest L index.
But a more efficient method: consider the relative order of the R’s and L’s only.