If frac{n!}{2!(n-2)!} and frac{n!}{4!(n-4)!} | vedclass

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(B) Given the ratio: $\frac{n!}{2!(n-2)!} : \frac{n!}{4!(n-4)!} = 2:1$.This can be written as: $\frac{n!}{2!(n-2)!} 	imes \frac{4!(n-4)!}{n!} = \frac

Vedclass · Accepted Answer

$5$

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(B) Given the ratio: $\frac{n!}{2!(n-2)!} : \frac{n!}{4!(n-4)!} = 2:1$.This can be written as: $\frac{n!}{2!(n-2)!} 	imes \frac{4!(n-4)!}{n!} = \frac{2}{1}$.Canceling $n!$ from the numerator and denominator: $\frac{4!(n-4)!}{2!(n-2)!} = 2$.Expanding the factorials: $\frac{4 	imes 3 	imes 2!}{2!} 	imes \frac{(n-4)!}{(n-2)(n-3)(n-4)!} = 2$.Simplifying the expression: $\frac{12}{(n-2)(n-3)} = 2$.Dividing both sides by $2$: $\frac{6}{(n-2)(n-3)} = 1$.So,$(n-2)(n-3) = 6$.Expanding the quadratic equation: $n^2 - 5n + 6 = 6$.$n^2 - 5n = 0$.$n(n-5) = 0$.Since $n$ must be at least $4$ for the term $(n-4)!$ to be defined,we have $n = 5$.

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$,find the value of $n$.

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