Permutations are all about arrangements — how many ways can we arrange things when the order matters. Choosing a president, then a vice-president, then a secretary from a group is a permutation problem because who gets which role matters. Let’s build this up step by step.
The Fundamental Counting Principle
Before we even touch formulas, we need this one idea:
If task 1 can be done in m ways, and task 2 can be done in n ways, then both tasks together can be done in m × n ways.
In simple language: multiply the choices at each step.
Example: We have 3 shirts and 4 pants. Total outfits = 3 × 4 = 12.
This extends to any number of tasks. If we have a 3-step process with 5, 3, and 2 choices at each step, total ways = 5 × 3 × 2 = 30.
Factorial
The factorial of n (written n!) is the product of all positive integers from 1 to n:
- 0! = 1 (by definition — this trips people up, just memorize it)
- 1! = 1
- 2! = 2
- 3! = 6
- 4! = 24
- 5! = 120
- 6! = 720
- 7! = 5040
- 8! = 40320
- 9! = 362880
- 10! = 3628800
Memorize up to 7! at least. These come up constantly.
Useful property: n! = n × (n-1)!. So 8! = 8 × 7! = 8 × 5040 = 40320.
Key Formulas
nPr — The Permutation Formula
nPr = n! / (n-r)! gives us the number of ways to arrange r items chosen from n distinct items.
- nPn = n! (arrange all n items — the (n-n)! = 0! = 1 in the denominator)
- nP1 = n (just picking 1 item from n — n ways)
- nP0 = 1 (doing nothing — there’s exactly 1 way to do nothing)
Example: How many 3-digit numbers can be formed from digits 1-5 (no repetition)?
5P3 = 5!/(5-3)! = 120/2 = 60
Think of it as: 5 choices for first digit × 4 for second × 3 for third = 60.
Arrangements with Repeated Items
When some items are identical, we divide by the factorial of each group of repeats:
Arrangements = n! / (p! × q! × r!…)
where p, q, r… are the frequencies of the repeated items.
Why? Because swapping identical items doesn’t create a new arrangement. If we have 3 identical A’s, those 3! = 6 swaps among themselves are all the same arrangement, so we divide by 3!.
The MISSISSIPPI Problem
This is the classic example. How many ways can we arrange the letters of MISSISSIPPI?
Letters: M(1), I(4), S(4), P(2) — total 11 letters
Arrangements = 11! / (1! × 4! × 4! × 2!) = 39916800 / (1 × 24 × 24 × 2) = 39916800 / 1152 = 34650
Circular Permutations
When we arrange items in a circle, we fix one item’s position (since rotations of the same arrangement are identical) and arrange the rest.
Circular permutations of n items = (n-1)!
In simple language: in a line, ABCD and BCDA are different. In a circle, they’re the same (just rotated). So we divide by n, giving n!/n = (n-1)!.
Necklace/Bracelet Problem
For a necklace or bracelet, clockwise and counterclockwise arrangements look the same (we can flip it). So we divide by 2 as well:
Necklace arrangements = (n-1)! / 2
Restricted Permutations
”Always Together” Problems
When certain items must always be next to each other:
- Bundle them into a single unit (treat the group as one item)
- Arrange the units (including the bundle)
- Arrange the items within the bundle
- Multiply the results
”Never Together” Problems
When certain items must never be adjacent:
Never together = Total arrangements − Always together
Or use the gap method: arrange the other items first, then place the restricted items in the gaps between them.
Fixed Position Problems
- Starts with / ends with a specific item: Fix that item, arrange the rest
- Even/odd positions only: Count the available positions, fill them, then fill the rest
Shortcut Methods and Tricks
Trick 1: Building numbers step by step For “how many n-digit numbers” problems, think step by step. For a 4-digit number:
- First digit: can’t be 0 (so count carefully)
- Remaining digits: fill one by one based on constraints
Trick 2: At least one / at most one “At least one” = Total - None “At most one” = None + Exactly one
Trick 3: Odd/Even number formations If the number must be even, fix the last digit first (it must be even), then fill the rest. If the number must be odd, fix the last digit first (it must be odd).
Trick 4: Division shortcut When calculating nPr, we don’t need the full factorial. 10P3 = 10 × 9 × 8 = 720. Just multiply from n going down, r times.
Worked Examples
Example 1: Basic Permutation
How many 4-letter words (with or without meaning) can be formed from the letters of REACT, without repeating any letter?
We have 5 distinct letters and need to arrange 4 of them.
5P4 = 5 × 4 × 3 × 2 = 120 words
Example 2: Number Formation
How many 3-digit even numbers can be formed using digits 1, 2, 3, 4, 5 (no repetition)?
For an even number, the last digit must be even: 2 or 4 → 2 choices First digit: any of the remaining 4 digits → 4 choices Second digit: any of the remaining 3 digits → 3 choices
Total = 4 × 3 × 2 = 24 numbers
Key point: We filled the most restricted position (last digit) first. Always handle constraints first!
Example 3: Word Arrangement with Repeats
How many ways can the letters of COMMITTEE be arranged?
Letters: C(1), O(1), M(2), I(1), T(2), E(2) — total 9 letters
Arrangements = 9! / (2! × 2! × 2!) = 362880 / 8 = 45360
Example 4: Always Together
In how many ways can the letters of GARDEN be arranged so that the vowels (A, E) are always together?
Treat the vowels as one unit: [AE], G, R, D, N → 5 units Arrange 5 units = 5! = 120 Arrange vowels within the unit = 2! = 2
Total = 120 × 2 = 240 ways
Example 5: Circular Permutation
In how many ways can 8 people sit around a circular table?
Circular permutation = (8-1)! = 7! = 5040 ways
If the table is replaced by a bracelet of 8 beads: 7!/2 = 5040/2 = 2520 arrangements.
Common Exam Variations
- Number formation with/without repetition, with constraints (even, odd, greater than X, divisible by 5)
- Word arrangements from words like INDEPENDENCE, MATHEMATICS, PERMUTATION (repeating letters)
- Circular seating — sometimes with constraints like “A and B must sit together” or “A and B must not sit adjacent”
- Books on a shelf — sometimes with “all math books together” or “no two science books adjacent”
- Flag signals — using colored flags in different arrangements
Practice Problems
Q1: How many 4-digit numbers greater than 5000 can be formed using digits 0, 1, 3, 5, 7 (no repetition)?
Q2: In how many ways can the letters of BANANA be arranged?
Q3: 6 people are to sit around a circular table. In how many ways can they sit such that two specific people (A and B) always sit next to each other?
Answers
A1: First digit must be 5 or 7 (to be > 5000) → 2 choices. Remaining 3 positions from 4 remaining digits = 4 × 3 × 2 = 24. Total = 2 × 24 = 48 numbers. But wait — check if any start with 0 in remaining positions. Since 0 isn’t in the first position, all are valid 4-digit numbers. Answer: 48.
A2: Letters: B(1), A(3), N(2) — total 6 letters. Arrangements = 6! / (3! × 2!) = 720 / 12 = 60.
A3: Treat A and B as one unit → 5 units around a circle = (5-1)! = 24. A and B can swap within the unit = 2! = 2. Total = 24 × 2 = 48 ways.