# nPr Formula — Permutations, Derivation, Examples

TL;DR

The nPr formula counts how many ordered arrangements of $r$ objects you can make from $n$ distinct objects: $^nP_r = \frac{n!}{(n-r)!}$. This article derives that formula from the counting principle, defines every symbol, shows where order makes a permutation different from a combination, and works six examples from a simple line-up to a locked-position arrangement.

## What Is the nPr Formula?

A permutation is an ordered selection. The nPr formula gives the number of ways to **pick and arrange** $r$ items from $n$ distinct items, where rearranging the chosen items counts as a new result.

$$\boxed{^nP_r = \frac{n!}{(n-r)!}}$$

| Symbol | Meaning | Notes |
| --- | --- | --- |
| $n$ | Total number of distinct objects available | A positive integer |
| $r$ | Number of objects selected and arranged | $0 \le r \le n$ |
| $n!$ | $n$ factorial — the product $n(n-1)(n-2)\cdots 1$, a [shrinking sequence](/content/math/algebra/sequences/index.html) of factors | $0! = 1$ by definition |
| $^nP_r$ | Number of ordered arrangements | Also written $P(n,r)$ or $_nP_r$ |

The condition that the $n$ objects are **distinct** matters — the formula assumes no repeats among the items, and that an item, once placed, is not used again. Permutations are the ordered half of combinatorics; the unordered half is counted by nCr, the combination formula.

## How Is the nPr Formula Derived?

The derivation comes straight from the **fundamental counting principle**: if one choice has $x$ options and the next has $y$, the two together have $x \times y$ outcomes. Filling $r$ ordered positions from $n$ objects works the same way.

- The **first** position can be filled $n$ ways.
- The **second** position then has $n - 1$ objects left.
- The **third** has $n - 2$, and so on.
- The **$r$-th** (last) position has $n - r + 1$ objects left.

Multiply the choices:

$$^nP_r = n(n-1)(n-2)\cdots(n-r+1).$$

That product is the top of a factorial cut short after $r$ terms. To write it cleanly, multiply and divide by $(n-r)!$:

$$^nP_r = \frac{n(n-1)\cdots(n-r+1)\cdot(n-r)!}{(n-r)!} = \frac{n!}{(n-r)!}.$$

The $(n-r)!$ in the denominator is exactly the part of $n!$ you do not use — the objects you never placed. That is the whole formula, and it is why $^nP_n = \frac{n!}{0!} = n!$: arranging all $n$ objects uses every factor.

## Examples of the nPr Formula

### Example 1

**Evaluate $^{10}P_4$.**

$$^{10}P_4 = \frac{10!}{(10-4)!} = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040.$$  
**Final answer:** $5040$.

### Example 2

**How many 3-letter arrangements can be made from the letters of the word MATH (no repeats)?**

**Wrong attempt.** A student thinks "choose 3 letters from 4" and computes $\frac{4!}{3!,1!} = 4$ using the combination formula. Check it against a small case: from M, A, T the arrangements MAT and TAM are clearly different words, yet the combination count treats them as one. Order matters here, so a combination undercounts.

**Correct.** Use the permutation formula:

$$^4P_3 = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 24.$$  
**Final answer:** $24$ arrangements.

### Example 3

**In how many ways can 8 runners finish 1st, 2nd, and 3rd in a race?**

Three ordered positions from eight runners:

$$^8P_3 = \frac{8!}{5!} = 8 \times 7 \times 6 = 336.$$  
**Final answer:** $336$ ways.

### Example 4

**How many ways can 6 books be arranged on a shelf?**

All six are arranged, so $r = n = 6$:

$$^6P_6 = \frac{6!}{0!} = 720.$$  
**Final answer:** $720$.

### Example 5

**How many 4-digit numbers can be formed from the digits 1–9 with no digit repeated?**

Nine distinct digits, four ordered positions:

$$^9P_4 = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024.$$  
**Final answer:** $3024$.

### Example 6

**Five people sit in a row of five seats, but two specific friends must sit at the two ends. How many arrangements are possible?**

Fix the two friends at the ends: they can swap, giving $^2P_2 = 2$ ways. The remaining three people fill the three middle seats in $^3P_3 = 6$ ways. By the counting principle, total $= 2 \times 6 = 12$.

**Final answer:** $12$ arrangements.

## Why Permutations Matter — From Passwords to PIN Pads

Counting ordered arrangements was forced into existence by problems where sequence is the answer, not a detail.

- **Security.** The number of possible PINs, passwords, and lock combinations is a permutation count — a 4-digit PIN with no repeats has $^{10}P_4 = 5040$ possibilities, which is exactly why short codes are weak.
- **Scheduling and rankings.** Seating charts, tournament brackets, race results, playlist orders — anything where "who is first" changes the outcome — is permutation territory.
- **Genetics and biology.** The order of bases along a DNA strand is a permutation problem; sequence determines the gene.
- **Cryptography.** Modern ciphers rest on how astronomically many ways symbols can be ordered, so that guessing the right order is infeasible.

The destination, in a later probability chapter, is computing the _odds_ of a specific arrangement: once you can count the arrangements, the probability of any one of them is one divided by that count. Counting cleanly is also what makes other reference formulas tractable — the same factorial machinery sits inside the sum of arithmetic sequence formula when you count terms before adding them.

## Tripping Points to Avoid

### Mistake 1: Using nPr when order does not matter

**Where it slips in:** Problems about committees, teams, handshakes, or lottery picks — where rearranging the chosen people changes nothing.

**Don't do this:** Apply $^nP_r$ to "choose 3 students for a committee from 10." A committee of {A, B, C} is the same committee as {C, B, A}.

**The correct way:** When order is irrelevant, use the combination formula $^nC_r = \frac{n!}{r!,(n-r)!}$.

### Mistake 2: Mishandling $0!$

**Where it slips in:** Cases where $r = n$, so the denominator is $0!$.

**Don't do this:** Treat $0!$ as $0$, which would make $^nP_n$ undefined.

**The correct way:** $0! = 1$ by definition. So $^nP_n = \frac{n!}{0!} = n!$, which matches the count of arranging all $n$ objects.

### Mistake 3: Forgetting the "distinct objects, no repetition" assumption

**Where it slips in:** Words with repeated letters (like BALLOON) or problems allowing repeats.

**Don't do this:** Apply plain $^nP_r$ to arrange the letters of a word with duplicates.

**The correct way:** When items repeat, divide by the factorial of each repeat count, or use the with-repetition rule $n^r$ when repeats are allowed. The bare nPr formula assumes all $n$ items are different and none is reused.

### Mistake 4: Subtracting in the wrong place

**Where it slips in:** Plugging into $\frac{n!}{(n-r)!}$ in a hurry.

**Don't do this:** Compute $\frac{(n-r)!}{n!}$ or $\frac{n!}{r!}$ by misremembering which factorial sits where.

**The correct way:** The bigger factorial $n!$ is on top; the denominator is $(n-r)!$.

## Conclusion

- The **nPr formula** counts ordered arrangements of $r$ from $n$ distinct objects: $^nP_r = \frac{n!}{(n-r)!}$.
- It is derived from the counting principle: $n$ choices for the first place, $n-1$ for the second, down to $n-r+1$ for the last.
- Use nPr when order matters and nCr when it does not — swapping two chosen items being a new result is the test.
- The most common mistake is using nPr for an unordered selection like a committee.
- Permutations underpin passwords, schedules, rankings, and the probability of a specific arrangement.

## Practice These Three Before Moving On

1. Evaluate $^7P_2$.
2. How many ways can 5 students be seated in 3 chairs?
3. How many 5-letter arrangements use the distinct letters of PRIME?

Answer to Question 1: $42$. Solve Questions 2 and 3 with $\frac{n!}{(n-r)!}$; if you are unsure whether order matters, return to Mistake 1.