Question

If $^{5}P_r = {}^{6}P_{r-1}$, then the value of $r$ is

• A. $r = 1$
• B. $r = 5$
• C. $r = 3$
• D. $r = 2$
• E. $r = 4$

Hint:

In permutation equations, always reduce factorial expressions step-by-step and cancel common terms early to simplify quickly.

Answer 1

The Correct Option is D

Concept:
The number of permutations of $n$ objects taken $r$ at a time is: \[ {}^nP_r = \frac{n!}{(n-r)!} \]

Step 1: Write both sides using permutation formula.
\[ {}^5P_r = \frac{5!}{(5-r)!} \] \[ {}^6P_{r-1} = \frac{6!}{(6-(r-1))!} = \frac{6!}{(7-r)!} \] So the equation becomes: \[ \frac{5!}{(5-r)!} = \frac{6!}{(7-r)!} \]

Step 2: Simplify factorials.
\[ 6! = 6 \cdot 5! \] Substitute: \[ \frac{5!}{(5-r)!} = \frac{6 \cdot 5!}{(7-r)!} \] Cancel $5!$ from both sides: \[ \frac{1}{(5-r)!} = \frac{6}{(7-r)!} \]

Step 3: Express $(7-r)!$ in terms of $(5-r)!$.
\[ (7-r)! = (7-r)(6-r)(5-r)! \] Substitute: \[ \frac{1}{(5-r)!} = \frac{6}{(7-r)(6-r)(5-r)!} \] Cancel $(5-r)!$: \[ 1 = \frac{6}{(7-r)(6-r)} \]

Step 4: Solve equation.
\[ (7-r)(6-r) = 6 \] Expand: \[ 42 - 13r + r^2 = 6 \] \[ r^2 - 13r + 36 = 0 \]

Step 5: Solve quadratic.
\[ r^2 - 13r + 36 = 0 \] \[ (r-9)(r-4) = 0 \] \[ r = 9 \text{or} r = 4 \]

Step 6: Check validity.
Since $r \leq 5$ (from ${}^5P_r$), valid solution is: \[ r = 4 \]
Final Answer: \[ \boxed{r = 4} \]