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Home » Let \( r = racab = rac2xy \), then: - Dachbleche24

Let \( r = racab = rac2xy \), then: - Dachbleche24

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 We proceed by counting the number of valid relative orders of roses (R₁, R₂) and lilies (L₁, L₂), treating them as distinguishable only by type, with the condition that all R's precede all L's.

📌 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?

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Total: 720
Hours: 6 Heads per hour? No: plan is 720 in 6 hours → 120/hr.
After 2 hours: 240
Third hour: +50 more than second hour → but second hour: same rate → 1...
So third hour: 120 + 50 = 170.
Total after 3 hours: 240 + 170 = 410
Remaining: 720 - 410 = 310
Last 3 hours: 310 ÷ 3 ≈ 103.333 — contradiction.
Ah! Here’s the error.
The instruction says: in the next hour walks twice as fast — so third...

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