Find $n$ if $^{n-1}P_3 : {}^nP_4 = 1 : 9$

> [!TLDR] Simplifying the ratio $^{n-1}P_3:{}^{n}P_4$ reduces to $1/n$, so setting $1/n=1/9$ gives $n=9$. > [!FORMULA] $^{n}P_{r}=\dfrac{n!}{(n-r)!}$, where $n$ is the total number of distinct items and $r$ is the number arranged. Here we use $^{n-1}P_3=\dfrac{(n-1)!}{(n-4)!}$ and $^{n}P_4=\dfrac{n!}{(n-4)!}$. > > 1. Write the given ratio: $\dfrac{^{n-1}P_3}{^{n}P_4}=\dfrac{1}{9}$. > > 2. Expand each permutation: $^{n-1}P_3=\dfrac{(n-1)!}{(n-1-3)!}=\dfrac{(n-1)!}{(n-4)!}$ and $^{n}P_4=\dfrac{n!}{(n-4)!}$. > > 3. Form the ratio: $\dfrac{^{n-1}P_3}{^{n}P_4}=\dfrac{(n-1)!/(n-4)!}{n!/(n-4)!}=\dfrac{(n-1)!}{n!}$. > > 4. Simplify using $n!=n\times(n-1)!$: $\dfrac{(n-1)!}{n!}=\dfrac{(n-1)!}{n\times(n-1)!}=\dfrac{1}{n}$. > > 5. Set this equal to the given ratio: $\dfrac{1}{n}=\dfrac{1}{9}$. > > 6. Solve: $n=9$. > > 7. Domain check: for $^{n-1}P_3$ and $^{n}P_4$ to be defined we need $n-1\ge3$ and $n\ge4$, i.e. $n\ge4$; $n=9$ satisfies this. > > 8. Self-check: substitute back — $^{8}P_3=8\times7\times6=336$ and $^{9}P_4=9\times8\times7\times6=3024$; ratio $=336/3024=1/9$ ✓, confirming $n=9$. > [!ANSWER] $n=9$.

Find n if ⁿ⁻¹P₃ : ⁿP₄ = 1 : 9

Exercise 6.3 · Q1

n

Sikkim CbseNCERTSubjective· 2mImportance★★★★★est

33% · 41/126 Questions

✓ Free question

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.

Simplifying the ratio $^{n-1}P_3:{}^{n}P_4$ reduces to $1/n$ , so setting $1/n=1/9$ gives $n=9$ .

[!FORMULA] $^{n}P_{r}=\dfrac{n!}{(n-r)!}$ , where $n$ is the total number of distinct items and $r$ is the number arranged. Here we use $^{n-1}P_3=\dfrac{(n-1)!}{(n-4)!}$ and $^{n}P_4=\dfrac{n!}{(n-4)!}$ .

Write the given ratio: $\dfrac{^{n-1}P_3}{^{n}P_4}=\dfrac{1}{9}$ .
Expand each permutation: $^{n-1}P_3=\dfrac{(n-1)!}{(n-1-3)!}=\dfrac{(n-1)!}{(n-4)!}$ and $^{n}P_4=\dfrac{n!}{(n-4)!}$ .
Form the ratio: $\dfrac{^{n-1}P_3}{^{n}P_4}=\dfrac{(n-1)!/(n-4)!}{n!/(n-4)!}=\dfrac{(n-1)!}{n!}$ .
Simplify using $n!=n\times(n-1)!$ : $\dfrac{(n-1)!}{n!}=\dfrac{(n-1)!}{n\times(n-1)!}=\dfrac{1}{n}$ .
Set this equal to the given ratio: $\dfrac{1}{n}=\dfrac{1}{9}$ .
Solve: $n=9$ .
Domain check: for $^{n-1}P_3$ and $^{n}P_4$ to be defined we need $n-1\ge3$ and $n\ge4$ , i.e. $n\ge4$ ; $n=9$ satisfies this.
Self-check: substitute back — $^{8}P_3=8\times7\times6=336$ and $^{9}P_4=9\times8\times7\times6=3024$ ; ratio $=336/3024=1/9$ ✓, confirming $n=9$ .

✓Final answer

n=9

←Previous question · 6.4Example 30 — How many arrangements of the letters of word SOCIOLOGICAL are there if…→Next question · 6.4Q2 — Find r if (i) ⁹Pr = 3024 (ii) ⁵Pr = 2⁶Pr-1

Q.Find nnn if n−1P3:nP4=1:9^{n-1}P_3 : {}^nP_4 = 1 : 9n−1P3​:nP4​=1:9

🎓Unlock everything free for 14 days

Q.Find $n$ if $^{n-1}P_3 : {}^nP_4 = 1 : 9$

Unlock everything free for 14 days