Find $r$ if (i) $^5P_r = 2\ ^6P_{r-1}$ (ii) $5\ ^4P_r = 6\ ^5P_{r-1}$

> [!TLDR] Write each permutation as a factorial ratio, cross-multiply, simplify to a quadratic in $r$, and keep the root consistent with the domain of the permutations. > [!FORMULA] > $^{n}P_r = \dfrac{n!}{(n-r)!}$, defined only for $0\le r\le n$. **(i)** $^{5}P_r = 2\,{}^{6}P_{r-1}$. 1. $^{5}P_r = \dfrac{5!}{(5-r)!}$ and $^{6}P_{r-1} = \dfrac{6!}{(6-(r-1))!} = \dfrac{6!}{(7-r)!}$. 2. Equation: $\dfrac{5!}{(5-r)!} = \dfrac{2\cdot 6!}{(7-r)!}$. 3. Since $(7-r)! = (7-r)(6-r)(5-r)!$, substitute: $\dfrac{5!}{(5-r)!} = \dfrac{2\cdot 6!}{(7-r)(6-r)(5-r)!}$. 4. Cancel $(5-r)!$ from both sides: $5! = \dfrac{2\cdot 6!}{(7-r)(6-r)} \Rightarrow (7-r)(6-r) = \dfrac{2\cdot 6!}{5!} = 2\times 6 = 12$. 5. Expand: $42 - 13r + r^2 = 12 \Rightarrow r^2 - 13r + 30 = 0 \Rightarrow (r-3)(r-10)=0 \Rightarrow r=3$ or $r=10$. 6. Domain check: $^{5}P_r$ needs $r\le 5$, so $r=10$ is rejected; $r=3$ remains. 7. Verify: $^{5}P_3 = 5\times4\times3=60$; $^{6}P_2=6\times5=30$; $2\times30=60$ ✓. **(ii)** $5\,{}^{4}P_r = 6\,{}^{5}P_{r-1}$. 8. $^{4}P_r = \dfrac{4!}{(4-r)!}$ and $^{5}P_{r-1} = \dfrac{5!}{(6-r)!}$. 9. Equation: $5\cdot\dfrac{4!}{(4-r)!} = 6\cdot\dfrac{5!}{(6-r)!}$. 10. Since $5! = 5\cdot4!$ and $(6-r)! = (6-r)(5-r)(4-r)!$: $\dfrac{5\cdot4!}{(4-r)!} = \dfrac{6\cdot5\cdot4!}{(6-r)(5-r)(4-r)!}$. 11. Cancel $4!/(4-r)!$: $5 = \dfrac{30}{(6-r)(5-r)} \Rightarrow (6-r)(5-r) = 6$. 12. Expand: $30 - 11r + r^2 = 6 \Rightarrow r^2 - 11r + 24 = 0 \Rightarrow (r-3)(r-8)=0 \Rightarrow r=3$ or $r=8$. 13. Domain check: $^{4}P_r$ needs $r\le4$, so $r=8$ is rejected; $r=3$ remains. 14. Verify: $^{4}P_3=4\times3\times2=24$, $5\times24=120$; $^{5}P_2=5\times4=20$, $6\times20=120$ ✓. > [!ANSWER] (i) $r=3$ (ii) $r=3$

Find r if (i) ⁵Pr = 2⁶Pr-1 (ii) 5⁴Pr = 6⁵Pr-1

Worked Examples · Example 18

r

(i)

^5P_r = 2\ ^6P_{r-1}

(ii)

5\ ^4P_r = 6\ ^5P_{r-1}

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Concept understanding — Fundamental Principles of Counting

The Fundamental Principle of Counting

Imagine you're getting dressed. You have 3 shirts (red, blue, green) and 2 pairs of pants (black, white). How many different outfits can you make?

You could list them: red-black, red-white, blue-black, blue-white, green-black, green-white. That's 6 outfits.

Notice something: 3 shirts × 2 pants = 6 outfits. That's not a coincidence — it's the entire idea.

The Intuition: A Fork in the Road

Think of each choice as a fork in a path. First you pick a shirt: 3 options. For each shirt, you then pick pants: 2 options. So the total number of paths through the whole process is 3 × 2 = 6.

This works no matter how many stages you add. If you also had 2 pairs of shoes, the total outfits become 3 × 2 × 2 = 12. Each new choice multiplies the total.

Note

The key insight: you multiply, not add. Addition would mean you choose either a shirt or pants, not both. Multiplication means you choose one of each, in sequence.

The Precise Statement

Fundamental Principle of Counting (Multiplication Principle)

If an event can happen in $m$ ways, and after it happens a second event can happen in $n$ ways, then the two events together can happen in $m \times n$ ways.

More generally: if a task consists of $k$ steps, and step 1 can be done in $n_1$ ways, step 2 in $n_2$ ways, ..., step $k$ in $n_k$ ways, then the total number of ways to complete the entire task is:

n_1 \times n_2 \times \cdots \times n_k

Watch out

This principle only works when the number of ways for each step does not depend on which choice was made earlier. If picking a red shirt somehow limits your pants options, you cannot simply multiply — you'd need to count more carefully.

A Simple Example

How many 2-digit numbers can you form using the digits 1, 2, 3, 4, 5 if repetition is allowed?

Step 1: Choose the tens digit — 5 options (1,2,3,4,5)
Step 2: Choose the units digit — 5 options (1,2,3,4,5)

Total = $5 \times 5 = 25$ numbers.

If repetition were not allowed:

Step 1: 5 options
Step 2: only 4 options (since one digit is already used)

Total = $5 \times 4 = 20$ numbers.

Tip

When repetition is not allowed, the number of options decreases by 1 at each step. When repetition is allowed, the number stays the same.

Why This Matters

This principle is the foundation of all counting problems in combinatorics. Permutations, combinations, arrangements — every one of them is just a special case of this multiplication idea, with some extra rules about repetition and order.

The rule is simple: when choices happen in sequence, multiply the number of options at each step.

Write each permutation as a factorial ratio, cross-multiply, simplify to a quadratic in $r$ , and keep the root consistent with the domain of the permutations.

[!FORMULA]

$^{n}P_r = \dfrac{n!}{(n-r)!}$ , defined only for $0\le r\le n$ .

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

$^{5}P_r = \dfrac{5!}{(5-r)!}$ and $^{6}P_{r-1} = \dfrac{6!}{(6-(r-1))!} = \dfrac{6!}{(7-r)!}$ .
Equation: $\dfrac{5!}{(5-r)!} = \dfrac{2\cdot 6!}{(7-r)!}$ .
Since $(7-r)! = (7-r)(6-r)(5-r)!$ , substitute: $\dfrac{5!}{(5-r)!} = \dfrac{2\cdot 6!}{(7-r)(6-r)(5-r)!}$ .
Cancel $(5-r)!$ from both sides: $5! = \dfrac{2\cdot 6!}{(7-r)(6-r)} \Rightarrow (7-r)(6-r) = \dfrac{2\cdot 6!}{5!} = 2\times 6 = 12$ .
Expand: $42 - 13r + r^2 = 12 \Rightarrow r^2 - 13r + 30 = 0 \Rightarrow (r-3)(r-10)=0 \Rightarrow r=3$ or $r=10$ .
Domain check: $^{5}P_r$ needs $r\le 5$ , so $r=10$ is rejected; $r=3$ remains.
Verify: $^{5}P_3 = 5\times4\times3=60$ ; $^{6}P_2=6\times5=30$ ; $2\times30=60$ ✓.

(ii) $5\,{}^{4}P_r = 6\,{}^{5}P_{r-1}$ .

8. $^{4}P_r = \dfrac{4!}{(4-r)!}$ and $^{5}P_{r-1} = \dfrac{5!}{(6-r)!}$ .

9. Equation: $5\cdot\dfrac{4!}{(4-r)!} = 6\cdot\dfrac{5!}{(6-r)!}$ .

10. Since $5! = 5\cdot4!$ and $(6-r)! = (6-r)(5-r)(4-r)!$ : $\dfrac{5\cdot4!}{(4-r)!} = \dfrac{6\cdot5\cdot4!}{(6-r)(5-r)(4-r)!}$ .

11. Cancel $4!/(4-r)!$ : $5 = \dfrac{30}{(6-r)(5-r)} \Rightarrow (6-r)(5-r) = 6$ .

12. Expand: $30 - 11r + r^2 = 6 \Rightarrow r^2 - 11r + 24 = 0 \Rightarrow (r-3)(r-8)=0 \Rightarrow r=3$ or $r=8$ .

13. Domain check: $^{4}P_r$ needs $r\le4$ , so $r=8$ is rejected; $r=3$ remains.

14. Verify: $^{4}P_3=4\times3\times2=24$ , $5\times24=120$ ; $^{5}P_2=5\times4=20$ , $6\times20=120$ ✓.

✓Final answer

r=3

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