Question

If \( {}^n P_r = {}^{n-1}P_r + x \cdot {}^{n-1}P_{r-1} \quad \forall n,r \in N \text{ and } r \le n \), then x = .

• A. (n+1)
• B. (r+1)
• C. n
• D. r

Hint:

Definition of permutation: \( {}^n P_r = \frac{n!}{(n-r)!} \).
Recurrence relation for permutations: \( {}^n P_r = {}^{n-1}P_r + r \cdot {}^{n-1}P_{r-1} \).
Another common one is \( {}^n P_r = n \cdot {}^{n-1}P_{r-1} \).
Substitute the factorial definitions and simplify algebraically.

Answer 1

The Correct Option is D

We know the formula for permutations: \( {}^n P_r = \frac{n!}{(n-r)!} \). The given recurrence relation is \( {}^n P_r = {}^{n-1}P_r + x \cdot {}^{n-1}P_{r-1} \). Let's expand the terms: \( {}^{n-1}P_r = \frac{(n-1)!}{((n-1)-r)!} = \frac{(n-1)!}{(n-r-1)!} \). \( {}^{n-1}P_{r-1} = \frac{(n-1)!}{((n-1)-(r-1))!} = \frac{(n-1)!}{(n-1-r+1)!} = \frac{(n-1)!}{(n-r)!} \). Substitute these into the equation: \( \frac{n!}{(n-r)!} = \frac{(n-1)!}{(n-r-1)!} + x \cdot \frac{(n-1)!}{(n-r)!} \) We can write \(n! = n \cdot (n-1)!\) and \((n-r)! = (n-r) \cdot (n-r-1)!\). \( \frac{n \cdot (n-1)!}{(n-r)(n-r-1)!} = \frac{(n-1)!}{(n-r-1)!} + x \cdot \frac{(n-1)!}{(n-r)(n-r-1)!} \) Assuming \((n-1)! \neq 0\) and \((n-r-1)! \neq 0\), divide by \(\frac{(n-1)!}{(n-r-1)!}\): \( \frac{n}{n-r} = 1 + x \cdot \frac{1}{n-r} \) Multiply by \((n-r)\): \( n = (n-r) + x \) \( n = n - r + x \) \( 0 = -r + x \) \( x = r \). This matches option (d). This recurrence relation \( {}^n P_r = {}^{n-1}P_r + r \cdot {}^{n-1}P_{r-1} \) is a known identity for permutations. \[ \boxed{r} \]