Permutations & Combinations
Grouping into pairs
Grade 11
Question:
<p>Sixteen players \(S_1, S_2, S_3, \ldots, S_{16}\) play in a tournament. Number of ways in which they can be grouped into eight pairs so that \(S_1\) and \(S_2\) are in different pairs, is</p>
<p>\(\dfrac{(14)!}{2^6 \cdot 6!}\)</p>
<p>\(\dfrac{(15)!}{2^7 \cdot 7!}\)</p>
<p>\(\dfrac{(14)!}{2^7 \cdot 6!}\)</p>
<p>\(\dfrac{(14)!}{2^6 \cdot 7!}\)</p>
Step-by-Step Solution
Key Concept: Subtract the cases where S₁ and S₂ are paired together from the total ways to partition 16 players into 8 pairs. The key is recognizing that pairing is an unordered partition, requiring division by 8! to account for pair permutations.
<p><strong>Step 1: Total ways to partition 16 players into 8 unordered pairs</strong></p><p>When forming pairs from 16 players, we arrange them in a line (16!) and divide by 2^8 (order within each pair doesn't matter) and 8! (order of pairs doesn't matter):</p><p>Total = 16!/(2^8 · 8!)</p><p><strong>Step 2: Ways where S₁ and S₂ are paired together</strong></p><p>If S₁ and S₂ must be together, treat them as one unit. We now have 15 units to partition into pairs (one pair is already fixed as {S₁, S₂}, leaving 14 other players to pair into 7 pairs):</p><p>Together = 14!/(2^7 · 7!)</p><p><strong>Step 3: Apply the subtraction principle</strong></p><p>Ways where S₁ and S₂ are in different pairs:</p><p>= 16!/(2^8 · 8!) − 14!/(2^7 · 7!)</p><p>= 16!/(2^8 · 8!) − 14!/(2^7 · 7!)</p><p>= [16 · 15 · 14!/(2^8 · 8 · 7!)] − [14!/(2^7 · 7!)]</p><p>= [14!/(2^7 · 7!)] · [16 · 15/(2 · 8) − 1]</p><p>= [14!/(2^7 · 7!)] · [240/16 − 1]</p><p>= [14!/(2^7 · 7!)] · [15 − 1]</p><p>= 14 · [14!/(2^7 · 7!)]</p><p>∴ Answer: <strong>D</strong></p>
Correct Answer: D