Simplify \(^{n + 1}C_{4} - ^{n - 1}C_{4}\) = \(\frac{(n + 1)!}{4! (n - 3)!} - \frac{(n - 1)!}{4! (n - 5)!}\) = \(\frac{(n + 1)(n)(n - 1)(n - 2)(n - 3)!}{4! ...

This question tests the ability to handle combinations symbolically. The trick is to write each \(^nC_r\) using \(^nC_r=\dfrac{n!}{r!\,(n-r)!}\), expand only as many factors from the top factorial as are needed to cancel the bottom factorial, and then factor out what the two terms have in common before simplifying. The working already shown is correct at every stage; below is the full derivation with each step made explicit.

Step 1: Write each term in factorial form. With \(r=4\), the denominator factorial for \(^{n+1}C_4\) is \((n+1-4)!=(n-3)!\), and for \(^{n-1}C_4\) it is \((n-1-4)!=(n-5)!\):

\[^{n+1}C_4=\frac{(n+1)!}{4!\,(n-3)!},\qquad ^{n-1}C_4=\frac{(n-1)!}{4!\,(n-5)!}.\]

Step 2: Peel off four factors from each numerator so the factorials cancel. Since \((n+1)!=(n+1)\,n\,(n-1)(n-2)\,(n-3)!\) and \((n-1)!=(n-1)(n-2)(n-3)(n-4)\,(n-5)!\):

\[^{n+1}C_4=\frac{(n+1)\,n\,(n-1)(n-2)\,(n-3)!}{4!\,(n-3)!}=\frac{(n+1)\,n\,(n-1)(n-2)}{4!},\]

\[^{n-1}C_4=\frac{(n-1)(n-2)(n-3)(n-4)\,(n-5)!}{4!\,(n-5)!}=\frac{(n-1)(n-2)(n-3)(n-4)}{4!}.\]

Step 3: Subtract and take out the common factor. Both expanded terms contain \(\dfrac{(n-1)(n-2)}{4!}\), so factor it out:

\[^{n+1}C_4-{}^{n-1}C_4=\frac{(n-1)(n-2)}{4!}\Big[\,n(n+1)-(n-3)(n-4)\,\Big].\]

Step 4: Expand and simplify the bracket. Multiply out each product carefully:

\[n(n+1)=n^2+n,\qquad (n-3)(n-4)=n^2-7n+12.\]

Subtracting, the \(n^2\) terms cancel:

\[n(n+1)-(n-3)(n-4)=(n^2+n)-(n^2-7n+12)=8n-12=4(2n-3).\]

Step 5: Substitute back and cancel. Using \(4!=24\):

\[^{n+1}C_4-{}^{n-1}C_4=\frac{(n-1)(n-2)\cdot 4(2n-3)}{24}=\frac{(n-1)(n-2)(2n-3)}{6}.\]

The factor \(4\) cancels the \(24\) down to \(6\), giving the fully simplified result:

\[\boxed{\;^{n+1}C_4-{}^{n-1}C_4=\frac{(n-1)(n-2)(2n-3)}{6}\;}\]

Where mistakes creep in: the most common slip is a sign error when expanding \(-(n-3)(n-4)\). It becomes \(-(n^2-7n+12)=-n^2+7n-12\), so the middle terms add to \(+8n\), not \(+6n\) or \(-8n\). A second frequent error is peeling off the wrong number of factors from the factorial: for \(^{n}C_4\) you always strip exactly four consecutive factors from the top, matching the four factors of \(4!\) implied in the denominator, then the remaining factorial cancels cleanly.

Examination reminder: when simplifying a difference of two combinations, expand each numerator only down to the largest common factorial, factor out the shared product before multiplying anything out, and reduce the numerical fraction (here \(4/24=1/6\)) at the very end.