Answer: \(n=40\).

Use

\({}^mC_r=\dfrac{m!}{r!(m-r)!}.\)

The equation is

\((n-4){}^{n+1}C_5={} ^{n+2}C_7.\)

Write each combination in factorial form:

\((n-4)\dfrac{(n+1)!}{5!(n-4)!}=\dfrac{(n+2)!}{7!(n-5)!}.\)

Since

\((n-4)!=(n-4)(n-5)!,\)

the factor \(n-4\) on the left cancels with the \((n-4)\) inside \((n-4)!\). Therefore the left-hand side becomes

\(\dfrac{(n+1)!}{5!(n-5)!}.\)

On the right-hand side, write

\((n+2)!=(n+2)(n+1)!\)

and

\(7!=7\cdot6\cdot5!=42\cdot5!.\)

So the right-hand side is

\(\dfrac{(n+2)(n+1)!}{42\cdot5!(n-5)!}.\)

Thus

\(\dfrac{(n+1)!}{5!(n-5)!}=\dfrac{(n+2)(n+1)!}{42\cdot5!(n-5)!}.\)

Cancel the common factor to get

\(1=\dfrac{n+2}{42}.\)

Therefore

\(n+2=42,\)

so

\(n=40.\)

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,

\(n=5.\)