Find $r$ if (i) ${}^{5}P_r = 2\, {}^{6}P_{r-1}$ (ii) ${}^{5}P_r = {}^{6}P_{r-1}$.

Question

Accepted Answer

> [!TLDR]
> The key idea is to use the permutation formula ${}^{n}P_r = \frac{n!}{(n-r)!}$ and simplify the factorial equation. For (i), $r = 3$; for (ii), $r = 2$.

Let’s start with the core concept. A permutation counts the number of ways to arrange $r$ distinct objects chosen from $n$ distinct objects, where order matters. The formula is:

> [!FORMULA]
> $${}^{n}P_r = \frac{n!}{(n-r)!}$$

This works because we have $n$ choices for the first position, $n-1$ for the second, and so on, down to $n-r+1$ for the $r$-th position. The factorial form is compact and lets us solve equations like these.

Now, we’ll solve each part step by step.

### Part (i): ${}^{5}P_r = 2\, {}^{6}P_{r-1}$

1. **Write both sides using the formula.**  
   Left side: ${}^{5}P_r = \frac{5!}{(5-r)!}$  
   Right side: $2 \cdot {}^{6}P_{r-1} = 2 \cdot \frac{6!}{(6-(r-1))!} = 2 \cdot \frac{6!}{(7-r)!}$

2. **Set up the equation.**  
   $$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(7-r)!}$$

3. **Simplify the factorials.**  
   Notice $6! = 6 	imes 5!$. Substitute:  
   $$\frac{5!}{(5-r)!} = 2 \cdot \frac{6 	imes 5!}{(7-r)!}$$

Cancel $5!$ from both sides (it’s non-zero):  
   $$\frac{1}{(5-r)!} = \frac{12}{(7-r)!}$$

4. **Cross-multiply.**  
   $$(7-r)! = 12 \cdot (5-r)!$$

5. **Expand the larger factorial.**  
   $(7-r)! = (7-r)(6-r)(5-r)!$  
   So:  
   $$(7-r)(6-r)(5-r)! = 12 \cdot (5-r)!$$

Cancel $(5-r)!$ (again, non-zero for valid $r$):  
   $$(7-r)(6-r) = 12$$

6. **Solve the quadratic.**  
   Expand: $42 - 13r + r^2 = 12$  
   Rearrange: $r^2 - 13r + 30 = 0$  
   Factor: $(r-3)(r-10) = 0$  
   So $r = 3$ or $r = 10$.

7. **Check domain restrictions.**  
   For ${}^{5}P_r$, we need $0 \leq r \leq 5$. For ${}^{6}P_{r-1}$, we need $0 \leq r-1 \leq 6$, i.e., $1 \leq r \leq 7$.  
   $r = 3$ satisfies both. $r = 10$ is invalid (exceeds both limits).  
   Thus, only $r = 3$ works.

> [!WARNING]
> A common mistake is to forget the domain restrictions on $r$. Permutations are defined only when $

Q.Find $r$ if

Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Why "Without Repetition"?

The Counting Logic – Why Multiply?

What If You Only Arrange Some of Them?

Key Points to Remember

Common Mistake to Avoid …

Part (i): ${}^{5}P_r = 2\, {}^{6}P_{r-1}$

Part (ii): ${}^{5}P_r = {}^{6}P_{r-1}$

Q.Find rrr if

Part (i): 5Pr=2 6Pr−1{}^{5}P_r = 2\, {}^{6}P_{r-1}5Pr​=26Pr−1​

Part (ii): 5Pr=6Pr−1{}^{5}P_r = {}^{6}P_{r-1}5Pr​=6Pr−1​

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Q.Find $r$ if

Part (i): ${}^{5}P_r = 2\, {}^{6}P_{r-1}$

Part (ii): ${}^{5}P_r = {}^{6}P_{r-1}$

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