101st airborne

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.

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