If {}^n{P_4} : {}^n{P_5} = 1 : 2,then n = | vedclass

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(C) The formula for permutation is ${}^n{P_r} = \frac{n!}{(n-r)!}$.Given the ratio: $\frac{{}^n{P_4}}{{}^n{P_5}} = \frac{1}{2}$.Substituting the formu

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$6$

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(C) The formula for permutation is ${}^n{P_r} = \frac{n!}{(n-r)!}$.Given the ratio: $\frac{{}^n{P_4}}{{}^n{P_5}} = \frac{1}{2}$.Substituting the formula: $\frac{n!}{(n-4)!} \div \frac{n!}{(n-5)!} = \frac{1}{2}$.This simplifies to: $\frac{n!}{(n-4)!} 	imes \frac{(n-5)!}{n!} = \frac{1}{2}$.Canceling $n!$ from numerator and denominator: $\frac{(n-5)!}{(n-4)!} = \frac{1}{2}$.Since $(n-4)! = (n-4) 	imes (n-5)!$,we have: $\frac{(n-5)!}{(n-4)(n-5)!} = \frac{1}{2}$.$\frac{1}{n-4} = \frac{1}{2}$.Therefore,$n-4 = 2$,which gives $n = 6$.

If ${}^n{P_4} : {}^n{P_5} = 1 : 2$,then $n = $

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