Question: A stack of $ 6n $ cards is numbered from 1 to $ 6n $, and each card belongs to one of $ n $ groups of 6 consecutive numbers. In how many ways can a person select 2 cards such that they are from different groups?

["Why This Simple Card Puzzle Is Surprisingly Popular (and How Many Ways Are There?) \nA curious question often surfaces among numbers enthusiasts, mobile learners, and pattern seekers: How many ways can a person select 2 cards from a stack of $ 6n $ cards, where each card belongs to one of $ n $ groups of 6 consecutive numbers, such that the two cards come from different groups? Against the backdrop of growing interest in structured challenges, logic puzzles, and interactive games on mobile, this question reflects a common desire to understand combinatorics in relatable, real-world scenarios. People naturally explore how treats, probabilities, and groupings work—even with abstract setups like numbered card stacks.", "Why This Question Resonates Today \nAcross the US, curiosity about puzzles, group dynamics, and pattern recognition continues to rise. From casual puzzle apps to family-friendly educational games, people seek structured challenges that stimulate critical thinking without explicit content. This card question taps into that trend: it feels approachable, grounded, and intellectually satisfying. Its appeal lies in simplicity—many users can visualize and explore the logic behind the math—making it ideal for mobile-first content seeking organic dwell time.", "Breaking Down the Problem: How Many Pairs Across Groups? \nEach group contains 6 consecutive card numbers, grouped into $ n $ total sets. To form a pair from different groups, we avoid selecting two cards within the same 6-card set. In each group of 6, the number of ways to choose 2 cards is $ \binom{6}{2} = 15 $. But since we want only cross-group pairs, the total eligible pairs are calculated by first finding all possible 2-card combinations from the full stack, then subtracting pairs that fall within the same group.", "Total ways to pick any 2 cards from $ 6n $ holds $ \binom{6n}{2} = \frac{6n(6n - 1)}{2} $. \nFrom each group of 6, invalid pairs total $ \binom{6}{2} = 15 $, and with $ n $ groups, total intra-group pairs are $ 15n $. \nThus, the number of valid inter-group selections is: \n\[\n\ ext{Valid pairs} = \binom{6n}{2} - 15n = \frac{6n(6n - 1)}{2} - 15n = 18n^2 - 3n - 15n = 18n^2 - 18n\n\]", "So, the final formula is: \n**"]