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How do I solve 11Pn = 12P(n-1) ? Please suggest some method (other than hit & trial, of course). Problem from Challenges and Thrills of Pre-College Mathematics (Excercise 9.1 Q6).

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Use the formula for $^n P_r $ and we have $$\frac {11!}{(11-n)!} =\frac {12!}{(13-n)!} $$ $$\Rightarrow \frac {12!}{11!} =\frac {(13-n)!}{(11-n)!} =(12-n)(13-n) $$ Can you take it from here? (Note that $n <12$).