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📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 Solution:** The problem can be solved using combinatorics. The robot needs to make a total of \(m + n\) moves: \(m\) moves to the right and \(n\) moves up. The number of distinct paths is equivalent to the number of different ways to arrange these moves, which is given by the binomial coefficient:

📌 \binom{m+n}{m} = \frac{(m+n)!}{m! \, n!}

📌 This expression counts the number of ways to choose \(m\) right moves (or equivalently \(n\) up moves) out of \(m+n\) total moves. Therefore, the number of distinct paths is:

📌 \boxed{\frac{(m+n)!}{m! \, n!}}

📌 Question:** A historian of science is analyzing the growth of scientific publications over time. If the number of publications doubles every \(k\) years, and the initial count is \(P_0\), express the number of publications \(P(t)\) as a function of time \(t\), and find \(P(2k)\).

📌 Solution:** The problem describes exponential growth. The number of publications at time \(t\) can be modeled by the function:

📌 P(t) = P_0 \cdot 2^{t/k}

📌 This expression arises because the publications double every \(k\) years, implying that after \(k\) years, the count is \(2P_0\), after \(2k\) years, it is \(2^2 P_0\), and so on. Thus, the general formula for the number of publications at any time \(t\) is:

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