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📅 May 15, 2026 👤 scraface

May 15, 2026

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📌 \frac{5!}{3!} = \frac{120}{6} = 20

📌 Total number of distinct words:

📌 9! \times \frac{5!}{3!} = 362880 \times 20 = 7257600

📌 Thus, the total number of distinct words is $\boxed{7257600}$

📌 Question: Five integers are selected uniformly at random from the set $\{0, 1, 2, \ldots, 50\}$. Let $ z $ be the sum of the two largest selected integers. What is the probability that $ z \geq 100 $?

📌 Solution: We are selecting 5 distinct integers independently and uniformly at random from $ \{0, 1, 2, \ldots, 50\} $, and let $ z $ be the sum of the two largest. We want $ \mathbb{P}(z \geq 100) $.

📌 Since the selection is with replacement not allowed (standard in such order statistics unless specified), we assume sampling without replacement for modeling realistic distinct values, though the space is large. However, since 51 values and 5 draws, and exact computation is complex, we approximate via order statistics and symmetry.

📌 But for precision, consider the maximum possible sum of two largest: 50 + 49 = 99.

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