Answer: \(n=22\).

Use the formula

\({}^{n}P_r=\dfrac{n!}{(n-r)!}.\)

Then

\({}^{n+1}P_5=(n+1)n(n-1)(n-2)(n-3),\)

and

\({}^{n}P_3=n(n-1)(n-2).\)

The equation is

\(\dfrac{{}^{n+1}P_5}{437}={}^{n}P_3.\)

Substitute the permutation expressions:

\(\dfrac{(n+1)n(n-1)(n-2)(n-3)}{437}=n(n-1)(n-2).\)

Since \(n\) must be large enough for these permutations to exist, \(n(n-1)(n-2)\ne0\). Divide both sides by \(n(n-1)(n-2)\):

\((n+1)(n-3)=437.\)

Expand:

\(n^2-2n-3=437.\)

So

\(n^2-2n-440=0.\)

Factorise:

\((n-22)(n+20)=0.\)

Since \(n\) must be positive,

\(n=22.\)