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Home » Thus, the sum becomes a telescoping series: - Dyverse

Thus, the sum becomes a telescoping series: - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 5Question: A soil scientist wants to analyze 4 different soil samples for microbial diversity. If there are 10 distinct soil types and 3 types of chemical treatments, how many ways can the scientist select 4 samples with exactly 2 types of chemical treatments applied?

📌 Solution: First, choose 4 distinct soil types from 10: $\dbinom{10}{4}$. Then, select 2 chemical treatments from 3: $\dbinom{3}{2}$. Multiply these results: $\dbinom{10}{4} \times \dbinom{3}{2} = 210 \times 3 = 630$. The total number of combinations is $\boxed{630}$.

📌 Question: A science fiction writer designs a starship with 7 crew roles and 5 alien biome systems. How many ways can 3 crew roles and 2 biome systems be selected for a mission if each system requires a unique role?

📌 Solution: First, choose 3 crew roles from 7: $\dbinom{7}{3}$. Then, select 2 biome systems from 5: $\dbinom{5}{2}$. Since each system needs a unique role, multiply by the number of ways to assign roles to biomes: $\dbinom{7}{3} \times \dbinom{5}{2} \times 3! = 35 \times 10 \times 6 = 2100$. However, if roles are independent of biomes, the answer simplifies to $\dbinom{7}{3} \times \dbinom{5}{2} = 35 \times 10 = 350$. Clarifying the problem's constraints, the most logical interpretation is $\boxed{350}$.

📌 Question: A bioinformatics developer creates a tool to analyze 6 gene sequences and 4 protein markers. How many ways can 2 sequences and 2 markers be selected if one specific marker is mandatory for all selections?

📌 Solution: Since one marker is mandatory, choose 1 additional marker from the remaining 3: $\dbinom{3}{1}$. Then, select 2 sequences from 6: $\dbinom{6}{2}$. Multiply these: $\dbinom{6}{2} \times \dbinom{3}{1} = 15 \times 3 = 45$. The total number of valid combinations is $\boxed{45}$.Question: A tech startup founder has 8 potential investment opportunities and wants to diversify by selecting 5, but two specific opportunities are mutually exclusive. How many valid investment portfolios can be formed?

📌 Solution: First, calculate the total number of ways to choose 5 opportunities from 8 without restrictions: $\binom{8}{5} = 56$. Next, subtract the number of invalid portfolios that include both mutually exclusive opportunities. If both are selected, the remaining 3 are chosen from the other 6: $\binom{6}{3} = 20$. Thus, the valid portfolios are $56 - 20 = 36$. The final answer is $\boxed{36}$.

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