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Home » We use the Euclidean algorithm to find the greatest common factor (GCF) of 68 and 102. - go-checkin.com

We use the Euclidean algorithm to find the greatest common factor (GCF) of 68 and 102. - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 We are to compute the probability that in 4 independent choices among 4 options (say labeled A, B, C, D, with equal likelihood), exactly one option appears twice, and two other distinct options appear once each, with the fourth position filled by a fourth distinct option — but wait: this would require 4 distinct options total, and one repeated. Since only 4 positions exist, and we want exactly one option repeated, and the other two being different, the only valid pattern is: one option appears twice, two others appear once, and the fourth option appears zero times — but that’s only 4 choices total. So the pattern is: one option repeated twice, and two other distinct options appearing once each — that uses up 2 + 1 + 1 = 4 choices, with one option appearing twice and two others once each, and one option not used at all. So the multiset of choices is of the form {A, A, B, C}, where A, B, C are distinct, and D is unused.

📌 We compute the probability of such a multiplicity pattern.

📌 Total number of possible outcome sequences: $4^4 = 256$, since each of the 4 decisions has 4 choices.

📌 Now count favorable outcomes.

📌 Step 1: Choose the option that appears twice: $\binom{4}{1} = 4$ ways.

📌 Step 2: Choose two distinct options from the remaining 3 to appear once each: $\binom{3}{2} = 3$ ways.

📌 Step 3: Count the number of distinct arrangements of the multiset {A, A, B, C}. The number of distinct permutations is:

📌 \frac{4!}{2!1!1!} = 12

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