Permutations & Combinations
Card problems
Grade 11
Question:
<p>Two different packs of cards are shuffled together. Cards are dealt equally among 4 players, each getting 13 cards. In how many ways can a player get his cards if no two cards are from the same suit with the same denomination (i.e., two cards are identical), 13 cards are to be selected from 52 cards where each card is two in number?</p>
Step-by-Step Solution
Key Concept: Each of the 52 distinct card types (13 denominations × 4 suits) appears exactly twice in the combined deck. Selecting 13 cards means choosing which 13 types to pick, then deciding which copy (Pack 1 or Pack 2) of each type to take—giving ${}^{52}C_{13} \times 2^{13}$ ways.
<p><strong>Step 1:</strong> Understand the setup. Two complete packs shuffled together gives 104 cards total: 2 copies each of 52 distinct card types (13 denominations × 4 suits).</p><p><strong>Step 2:</strong> A player receives 13 cards. Since no two cards can be identical (same denomination AND suit), the player must get cards from exactly 13 different card types.</p><p><strong>Step 3:</strong> First, choose which 13 card types (out of 52) the player will receive: ${}^{52}C_{13}$ ways.</p><p><strong>Step 4:</strong> For each of the 13 chosen card types, the player must select 1 copy from the 2 available copies (one from each pack). This is a binary choice for each type: $2^{13}$ ways.</p><p><strong>Step 5:</strong> By the multiplication principle, the total number of ways is:</p><p style='text-align:center'>$${}^{52}C_{13} \times 2^{13} = \dfrac{52!}{13! \cdot 39!} \times 2^{13}$$</p><p><strong>∴ Answer:</strong> ${}^{52}C_{13} \times 2^{13}$</p>
Correct Answer: \({}^{52}C_{13} \times 2^{13} = \dfrac{52!}{13! \cdot 39!} \cdot 2^{13}\)