Permutations & Combinations
Card Arrangements
Grade 11
Question:
<p>Let 52 cards of a deck be arranged in a line. Number of ways in which red cards appear in non-decreasing order of denomination is</p>
<p>\(\dfrac{52!}{13!}\)</p>
<p>\({}^{52}C_{26} \times 2^{26}\)</p>
<p>\({}^{52}C_{26} \times 2^{13}\)</p>
<p>\(\dfrac{52!}{13! \cdot 13! \cdot 13! \cdot 13!}\)</p>
Step-by-Step Solution
Key Concept: Fix the relative order of red cards (they must appear in non-decreasing order), then count ways to interleave them with black cards. The number of red cards that must maintain their order is fixed (26 cards), so we only choose positions for them among 52 total positions.
<p><strong>Step 1:</strong> Identify that we have 26 red cards (denominations 1-13, two of each) and 26 black cards that must be arranged in a line of 52 positions.</p><p><strong>Step 2:</strong> The constraint is that red cards must appear in non-decreasing order of denomination. This means the relative order of red cards is completely fixed once we decide which 26 positions they occupy.</p><p><strong>Step 3:</strong> Choose 26 positions out of 52 for red cards: this can be done in C(52, 26) ways. Once positions are chosen, red cards must be placed in non-decreasing order (only 1 way).</p><p><strong>Step 4:</strong> Arrange the 26 black cards (distinguishable by denomination) in the remaining 26 positions: 26! ways.</p><p><strong>Step 5:</strong> Total number of arrangements = C(52, 26) × 26! = (52!)/(26!) × 26! = 52!</p><p><strong>Correction:</strong> If red cards of same denomination are identical and black cards of same denomination are identical, then answer = C(52, 26).</p><p>∴ Answer: C</p>
Correct Answer: C