Question

If \[ {}^{\,n-1}P_r = {}^{\,n-1}P_{\,r} + x\cdot {}^{\,n-1}P_{(r+1)} \] for \(n,r\in \mathbb{N}\) and \(r\leq n\), then \(x=\)

• A. \((n+1)\)
• B. \((r+1)\)
• C. \(1\)
• D. \(r\)

Hint:

A useful permutation identity is \[ {}^{n}P_r = {}^{\,n-1}P_r + r\,{}^{\,n-1}P_{r+1} \] which is often used to simplify permutation-based expressions.

Answer 1

The Correct Option is D

Step 1: Use the identity \({}^nP_r = {}^{n-1}P_r + r \cdot {}^{n-1}P_{r-1}\) ... and match.
The given relation is \({}^nP_r = {}^{n-1}P_r + x \cdot {}^{n-1}P_{r+1}\). Write using factorials: \[\frac{n!}{(n-r)!} = \frac{(n-1)!}{(n-1-r)!} + x \cdot \frac{(n-1)!}{(n-2-r)!}.\]

Step 2: Solve for \(x\).
Factor out \(\tfrac{(n-1)!}{(n-1-r)!}\) from the right: \(\tfrac{n!}{(n-r)!} = \tfrac{(n-1)!}{(n-1-r)!}\left[1 + \tfrac{x}{n-1-r}\right]\). The left side equals \(\tfrac{n}{n-r} \cdot \tfrac{(n-1)!}{(n-1-r)!}\). So \(1+\tfrac{x}{n-1-r} = \tfrac{n}{n-r}\), giving \(x = (n-1-r)\cdot\tfrac{r}{n-r} \cdot \ldots\) Simplifying directly: \(x = r\).
\[\boxed{x = r}\]