Solution: We need to compute the number of ways to choose 3 tokens from 8 and 2 signatures from 5. Since order does not matter in a packet, we use combinations.
["Understanding Combinatorial Selection: Choosing 3 Tokens from 8 and 2 Signatures from 5 Using Combinations", "When solving counting problems involving selections without regard to order — such as choosing tokens or signatures — combinations are the fundamental tool. This article explains how to compute the number of ways to choose 3 tokens from 8 and 2 signatures from 5, demonstrating why combinations, not permutations, are the correct approach.", "### Why Use Combinations?", "In mathematics, a combination calculates the number of ways to select items from a larger set where the order of selection does not matter. This is ideal for problems like choosing sets of tokens or signatures where only the group matters, not the sequence.", "For example, selecting tokens ABC from {A, B, C, D, E, F, G, H} yields the same set regardless of order — ABC = BCA = CAB. Thus, combinations give the correct count.", "### Step-by-step: Choosing 3 Tokens from 8", "To compute the number of ways to choose 3 tokens from 8, we apply the combination formula:", "[\nC(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}\n]", "Here,\n- ( n = 8 ) (total tokens)\n- ( k = 3 ) (tokens to choose)", "[\n\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} = \frac{8 \ imes 7 \ imes 6}{3 \ imes 2 \ imes 1} = \frac{336}{6} = 56\n]", "So, there are 56 distinct ways to choose 3 tokens from 8.", "### Step-by-step: Choosing 2 Signatures from 5", "Similarly, for choosing 2 signatures from 5:", "[\n\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!} = \frac{5 \ imes 4}{2 \ imes 1} = \frac{20}{2} = 10\n]", "There are 10 distinct ways to choose 2 signatures from 5.", "### Total Number of Packet Combinations", "Because token choices and signature choices are independent, we multiply the two results to get the total number of valid packet configurations:", "[\n\ ext{Total combinations} = \binom{8}{3} \ imes \binom{5}{2} = 56 \ imes 10 = 560\n]", "Thus, there are 560 unique packets possible by selecting 3 tokens from 8 and 2 signatures from 5.", "### Summary", "To count subsets where order does not matter:", "- Use combinations:\n [\n \binom{n}{k} = \frac{n!}{k!(n-k)!}\n ]", "- Multiply results when selections are independent.", "This approach applies across disciplines — from probability theory to database query optimization — wherever unordered selection matters.", "---", "Understanding combinations simplifies complex counting problems and ensures accurate results in combinatorics and applied mathematics."]
