Solution: The number of ways to choose 3 18th-century manuscripts is $inom{5}{3}$. Total ways to choose 3 manuscripts from 15 is $inom{15}{3}$. The probability is $ rac{inom{5}{3}}{inom{15}{3}} = rac{10}{455} = rac{2}{91}$. $oxed{\dfrac{2}{91}}$Question: If the combined population of two marine mammal species in a region is 10 and the sum of their squared populations is 50, find the sum of their cubes.

Solution: The number of ways to choose 3 18th-century manuscripts is $inom{5}{3}$. Total ways to choose 3 manuscripts from 15 is $inom{15}{3}$. The probability is $rac{inom{5}{3}}{inom{15}{3}} = rac{10}{455} = rac{2}{91}$. $oxed{\dfrac{2}{91}}$Question: If the combined population of two marine mammal species in a region is 10 and the sum of their squared populations is 50, find the sum of their cubes.

["Solution: Finding the Sum of Cubes from Population Data", "Understanding how fragmented data can reveal deeper patterns is crucial in fields like biology and ecology. In this article, we explore a conceptual problem inspired by marine mammal population modeling—demonstrating how combinatorial thinking supports analytical reasoning, while also solving a related algebraic challenge rooted in population values.", "---", "### Problem Statement", "Suppose two marine mammal species inhabit a region with a combined population of 10 individuals. The sum of the squares of their populations is 50. We are to find the sum of their cubes.", "Let the populations be $ x $ and $ y $. We are given:\n$$\nx + y = 10 \quad \ ext{(1)}\n$$\n$$\nx^2 + y^2 = 50 \quad \ ext{(2)}\n$$\nOur goal is to compute:\n$$\nx^3 + y^3\n$$", "---", "### Step 1: Use Identity for Sum of Cubes", "We apply the algebraic identity:\n$$\nx^3 + y^3 = (x + y)^3 - 3xy(x + y)\n$$", "We already know $ x + y = 10 $, so $ (x + y)^3 = 10^3 = 1000 $. The challenge is finding $ xy $.", "---", "### Step 2: Compute $ xy $ Using Given Data", "From equation (1), square both sides:\n$$\n(x + y)^2 = x^2 + 2xy + y^2\n$$\nSubstitute known values:\n$$\n10^2 = 50 + 2xy\n$$\n$$\n100 = 50 + 2xy \quad \Rightarrow \quad 2xy = 50 \quad \Rightarrow \quad xy = 25\n$$", "---", "### Step 3: Substitute into the Cubic Identity", "Now plug values into the sum of cubes formula:\n$$\nx^3 + y^3 = (10)^3 - 3(25)(10) = 1000 - 750 = 250\n$$", "---", "### Final Answer", "The sum of the cubes of the populations is:\n$$\n\boxed{250}\n$$", "---", "This elegant connection between algebraic identities and real-world ecological data highlights how mathematics enables precise modeling of complex biological systems. Just like calculating probabilities in manuscript selection, solving population puzzles requires logical consistency and careful manipulation of equations.", "Note: While the combinatorial example $ \dfrac{\binom{5}{3}}{\binom{15}{3}} = \dfrac{2}{91} $ matches its simplification to $ \dfrac{10}{455} $, the core mathematical reasoning—working through equations with symmetry, identities, and substitutions—transcends disciplines, from ancient manuscripts to modern marine biology."]

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