<p>Number of ways in which cards of each suite appear in increasing order of denomination is</p>
<p>\(\dfrac{52!}{13! \cdot 13! \cdot 13!}\)</p>
<p>\(\dfrac{52!}{13! \cdot 13! \cdot 13! \cdot 13!}\)</p>
<p>\({}^{52}C_{26} / 2^{13}\)</p>
<p>\(\dfrac{52!}{13! \cdot 13! \cdot 13! \cdot 14!}\)</p>
Step-by-Step Solution
Key Concept: When cards of each suit must appear in increasing order of denomination, we only need to choose which positions each suit occupies—the internal arrangement within each suit is fixed (automatically increasing). This reduces the problem to arranging 4 indistinguishable groups of 13 cards each into 52 positions.
<p><strong>Step 1:</strong> Recognize that each suit (13 cards) must appear in increasing order of denomination (A, 2, 3, ..., K). This means once we decide which 13 positions go to hearts, which 13 to diamonds, etc., the arrangement within each suit is completely determined—there's only 1 way to arrange each suit.</p><p><strong>Step 2:</strong> The problem reduces to: "In how many ways can we distribute 52 positions among 4 suits such that each suit gets exactly 13 positions?"</p><p><strong>Step 3:</strong> This is a multinomial coefficient problem. We need to choose 13 positions out of 52 for suit 1, then 13 out of remaining 39 for suit 2, then 13 out of remaining 26 for suit 3, and the last 13 go to suit 4.</p><p><strong>Step 4:</strong> Number of ways = $\frac{52!}{13! \times 13! \times 13! \times 13!}$</p><p><strong>Step 5:</strong> This can also be written as $\binom{52}{13,13,13,13} = \frac{52!}{(13!)^4}$</p><p>∴ Answer: B</p>
Correct Answer: B