\dbinom{7}{3} \times 2! \times 4! = 35 \times 2 \times 24 = \boxed{1680}

["Solving the Equation: \dbinom{7}{3} × 2! × 4! = 1680", "Calculating combinatorial expressions might seem complex at first, but breaking it down step-by-step reveals the elegance of factorials and binomial coefficients. In this article, we’ll explore how evaluating (\dbinom{7}{3} \ imes 2! \ imes 4! = 35 \ imes 2 \ imes 24 = 1680) delivers a powerful real-world insight in permutations and probability.", "---", "### Understanding the Components", "At the heart of this equation are three mathematical elements: the binomial coefficient (\dbinom{7}{3}), the factorial (2!), and the factorial (4!).", "- (\dbinom{7}{3}): This represents the number of ways to choose 3 items from 7 without regard to order. It’s the binomial coefficient, calculated as:\n [\n \dbinom{7}{3} = \frac{7!}{3!(7−3)!} = \frac{7!}{3!4!}\n ]\n- (2!): This is simply 2 × 1 = 2, the number of ways to arrange 2 distinct items.\n- (4!): This equals 4 × 3 × 2 × 1 = 24, representing the arrangements of 4 distinct items.", "---", "### Step-by-step Computation", "Start with the key expression:\n[\n\dbinom{7}{3} \ imes 2! \ imes 4!\n]", "First, substitute (\dbinom{7}{3} = \frac{7!}{3!4!}):\n[\n\frac{7!}{3!4!} \ imes 2! \ imes 4! = \frac{7!}{3!4!} \ imes 2! \ imes 4!\n]", "Notice (4!) cancels on numerator and denominator:\n[\n\frac{7!}{3!} \ imes 2! = \frac{7!}{3!} \ imes 2\n]", "Now compute (7! = 5040) and (3! = 6):\n[\n\frac{5040}{6} = 840\n]", "Then multiply by (2):\n[\n840 \ imes 2 = 1680\n]", "---", "### Why This Matters: Real-World Applications", "This formula arises in combinatorics when counting arrangements involving grouped options. For example, imagine choosing:", "- 3 team members out of 7 for a special task (ways = (\dbinom{7}{3} = 35)),\n- Arranging 2 alternative roles among them ((2! = 2) permutations),\n- Arranging the remaining 4 team members in a line ((4! = 24) permutations).", "Thus, total configurations are:\n[\n35 \ imes 2 \ imes 24 = 1680\n]", "Such calculations underpin probability problems, data analysis, and algorithmic complexity in computer science.", "---", "### Final Calculation Recap", "[\n\dbinom{7}{3} \ imes 2! \ imes 4! = 35 \ imes 2 \ imes 24 = \boxed{1680}\n]", "---", "### Conclusion", "The equation (\dbinom{7}{3} \ imes 2! \ imes 4! = 1680) is more than a number crunch—it’s a clean demonstration of how combinatorial mathematics organizes possibilities. Whether planning groups, scheduling events, or analyzing permutations, mastering such expressions empowers precise reasoning in countless fields.", "---", "Keywords: binomial coefficient, binomial coefficient calculation, factorial, permutations, combinations, (\dbinom{7}{3}), (2!), (4!), combinatorics, math tutorial, probability, arrangements, factorial calculation."]