Answer: \(n=5\).

The number of permutations of \(n\) items taken \(4\) at a time is

\({}^nP_4=n(n-1)(n-2)(n-3).\)

The number of permutations of \(2n\) items taken \(3\) at a time is

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

The question states that

\({}^nP_4=\dfrac16{}^{2n}P_3.\)

So

\(n(n-1)(n-2)(n-3)=\dfrac16(2n)(2n-1)(2n-2).\)

Since \(2n-2=2(n-1)\), the right-hand side becomes

\(\dfrac16(2n)(2n-1)\cdot2(n-1)=\dfrac23n(n-1)(2n-1).\)

Thus

\(n(n-1)(n-2)(n-3)=\dfrac23n(n-1)(2n-1).\)

For permutations taken \(4\) at a time, \(n\geqslant4\), so \(n(n-1)\ne0\). Divide by \(n(n-1)\):

\((n-2)(n-3)=\dfrac23(2n-1).\)

Multiplying by \(3\),

\(3(n-2)(n-3)=2(2n-1).\)

Expanding gives

\(3(n^2-5n+6)=4n-2.\)

Therefore

\(3n^2-19n+20=0,\)

as required.

Now factorise:

\(3n^2-19n+20=(3n-4)(n-5).\)

So \(n=\dfrac43\) or \(n=5\). Since \(n\) must be an integer number of items,

\(n=5.\)