Permutations & Combinations
Arrangements with Restrictions
Grade 11
Question:
<p>The number of ways of arranging <i>n</i> persons, if out of any two seats located symmetrically in the middle of the row at least one is empty is</p>
<p>\({}^{m/2}C_n(2^n) \cdot 1\)</p>
<p>\({}^{m/2}P_n\)</p>
<p>\(({}^{m/2}P_n)(2^n - 1)\)</p>
<p>\(({}^{m/2}P_n)(2^n)\)</p>
Step-by-Step Solution
Key Concept: Use complementary counting: total arrangements minus arrangements where both symmetric middle seats are occupied. The symmetric middle seats are positions (n/2) and (n/2 + 1) for even n.
<p><strong>Step 1:</strong> For n persons in a row, identify symmetric middle seats. For even n, these are positions n/2 and (n/2 + 1). For odd n, there is no pair of symmetric middle seats (only one center seat).</p><p><strong>Step 2:</strong> If n is odd: All n! arrangements are valid since there's no symmetric pair of seats. Answer: n!</p><p><strong>Step 3:</strong> If n is even: Use complementary counting.</p><p>• Total arrangements = n!</p><p>• Arrangements where BOTH symmetric middle seats are occupied = (arrangements with 2 specific persons in those 2 seats) × (arrangements of remaining persons) = 2! × (n-2)! = 2(n-2)!</p><p><strong>Step 4:</strong> Apply complement principle: Valid arrangements = n! - 2(n-2)! = (n-2)![n(n-1) - 2] = (n-2)!(n² - n - 2) = (n-2)!(n-2)(n+1)</p><p><strong>Step 5:</strong> Simplify: = (n-1)!(n+1) - 2(n-2)! = <strong>(n-2)!(n² - n - 2)</strong></p><p>For even n: <strong>(n-2)!(n-2)(n+1)</strong></p><p>∴ Answer: C</p>
Correct Answer: C