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Home » Can You Believe This UFC Start Time — Millions Didn’t See It Coming! 🔥 - ToelettAPP

Can You Believe This UFC Start Time — Millions Didn’t See It Coming! 🔥 - ToelettAPP

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Now compute $ 8104 \mod 8 $. Since 8100 is divisible by 8 (as $ 8100 \div 8 = 1012.5 $, but $ 8000 \div 8 = 1000 $, and $ 100 \div 8 = 12.5 $), we check only the last three digits:

📌 \div 8 = 13 \text{ exactly}, so } 8104 \equiv 0 \pmod{8}

📌 $$Question: How many ways are there to distribute 4 distinct chemical samples into 2 identical storage containers such that each container has at least one sample?

📌 Solution: The problem requires counting the number of ways to partition 4 distinct items into 2 non-empty identical subsets. This is given by the Stirling numbers of the second kind, $ S(4, 2) $. The formula for $ S(n, k) $ is $ S(n, k) = S(n-1, k-1) + k \cdot S(n-1, k) $. Using known values, $ S(4, 2) = 7 $. Thus, the number of ways is $ \boxed{7} $.

📌 Question: A philosopher examines 5 ethical guidelines. What is the probability that exactly 2 out of 3 randomly selected guidelines prioritize human welfare, if 3 guidelines are welfare-focused and 2 are not?

📌 Solution: The total number of ways to choose 3 guidelines from 5 is $ \binom{5}{3} = 10 $. The number of favorable outcomes (2 welfare, 1 non-welfare) is $ \binom{3}{2} \cdot \binom{2}{1} = 3 \cdot 2 = 6 $. The probability is $ \frac{6}{10} = \frac{3}{5} $. Thus, the probability is $ \boxed{\dfrac{3}{5}} $.

📌 Question: How many distinct 6-letter arrangements can be formed from the letters of "THEORETICAL" if the letters 'E' and 'A' must not be adjacent?

📌 Solution: First, calculate the total number of arrangements of the letters in "THEORETICAL". The word has 10 letters with repetitions: T (2), E (3), H (1), O (1), R (1), I (1), C (1), A (1). Total arrangements: $ \frac{10!}{2! \cdot 3!} = 302400 $. Next, subtract the arrangements where 'E' and 'A' are adjacent. Treat 'EA' or 'AE' as a single entity, reducing the problem to arranging 9 items (with T repeated twice and E reduced to 2). Number of such arrangements: $ 2 \cdot \frac{9!}{2! \cdot 2!} = 2 \cdot 362880 / 4 = 181440 $. The valid arrangements are $ 302400 - 181440 = 120960 $. Thus, the answer is $ \boxed{120960} $.Question: A 5 cm by 8 cm rectangle is inscribed in a circle. What is the number of centimeters in the circumference of the circle? Express your answer in terms of $\pi$.

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