If (\frac{n!}{(n-5)!}=55440) then what is the value of (n)?

Correct answer: (11)

The direct answer is option B, \(n=11\). Use the factorial cancellation rule: \(\frac{n!}{(n-5)!}=n(n-1)(n-2)(n-3)(n-4)\). We need five consecutive descending factors whose product is 55440. Test option B: \(11\cdot10\cdot9\cdot8\cdot7\). Stepwise, \(11\cdot10=110\), \(9\cdot8=72\), and \(110\cdot72\cdot7=7920\cdot7=55440\). Thus \(n=11\). Option A gives \(10\cdot9\cdot8\cdot7\cdot6=30240\), not the required value. Option C gives \(12\cdot11\cdot10\cdot9\cdot8=95040\). Option D gives an even larger product, \(13\cdot12\cdot11\cdot10\cdot9=154440\). Also, factorials require \(n\ge5\), which all choices satisfy. Memory cue: cancel factorials first, then count exactly five descending factors.