Permutations & Combinations
Card problems
Grade 11
Question:
<p>Two packs of 52 cards are shuffled together. The number of ways in which a man can be dealt 26 cards so that he does not get two cards of the same suit and same denomination is</p>
<p>\({}^{52}C_{26} \times 2^{26}\)</p>
<p>\({}^{104}C_{26}\)</p>
<p>\(2 \times {}^{52}C_{26}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use complementary counting: Total ways to choose 26 cards minus ways that include at least one pair of identical cards (same suit and denomination). Since there are 104 cards total with 2 identical copies of each of 52 cards, we must exclude arrangements containing duplicates.
<p><strong>Step 1:</strong> Recognize that we have 2 identical packs, so there are exactly 2 copies of each of the 52 distinct cards (same suit and denomination).</p><p><strong>Step 2:</strong> The constraint is: the man must get 26 cards such that he doesn't get both copies of any card. This means from each of the 52 card types, he gets at most 1 card.</p><p><strong>Step 3:</strong> To select 26 cards with at most 1 of each denomination-suit type: First choose which 26 card types (out of 52) to include: C(52,26) ways.</p><p><strong>Step 4:</strong> For each of the 26 selected card types, choose which copy (from the 2 available packs) to take: 2^26 ways.</p><p><strong>Step 5:</strong> Total number of ways = C(52,26) × 2^26</p><p>∴ Answer: A</p>
Correct Answer: A