A university professor is assigning 6 different research projects to 3 students such that each student receives exactly 2 projects. In how many distinct ways can the projects be distributed?
["A university professor is assigning 6 different research projects to 3 students such that each student receives exactly 2 projects. In how many distinct ways can the projects be distributed?", "Across U.S. universities, collaborative learning experiences are reshaping how instructors structure teamwork—especially in academic research. The scenario of a professor distributing six unique research projects among three students, with each receiving exactly two, surfaces repeatedly in faculty planning circles and instructional design research. This structure ensures balanced workloads and fosters peer learning, making it a common yet nuanced logistical challenge—one that holds quiet relevance for educators and students alike.", "But beyond classroom logistics, this assignment pattern raises a fascinating puzzle of combinatorics: how many distinct ways can six unique projects be split so that each of the three students receives exactly two? For users searching for clear, reliable answers—especially those interested in academic planning, educational strategy, or research facilitation—this question surfaces naturally within broader trends around collaborative learning models, workload equity, and project-based pedagogy.", "This isn’t just about counting arrangements—it’s about understanding coordination mechanics in educational systems, where fairness, efficiency, and intellectual engagement intersect. Let’s unpack the math behind this setup to reveal how distribution patterns form.", "---", "H3: The Math Simplified—Why It Still Matters", "Distributing six distinct research projects among three students, each getting exactly two, follows a structured combinatorial principle. At first glance, it appears as a simple division: \nFirst, select 2 projects from 6 for Student A: \n\[\n\binom{6}{2} = 15 \ ext{ ways}\n\] \nThen, choose 2 from the remaining 4 for Student B: \n\[\n\binom{4}{2} = 6 \ ext{ ways}\n\] \nStudent C automatically gets the last 2. \nMultiplying: \n\[\n15 \ imes 6 = 90\n\] \nBut this counts order: assigning specific groups to labeled students. Since the students’ identities are fixed (not interchangeable), we do not divide by permutations of the groups—this count of 90 is accurate for labeled recipients.", "A clearer combinatorial approach uses multinomial coefficients: \n\[\n\frac{6!}{2!2!2!} = \frac{720}{"]
