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Fundamentals of Business Mathematics and Statistics · Permutation and Combinations

Permutations of Distinct Objects: Formula nPr Explained

Updated 10 October 2026 · Fact-checked

A permutation is an arrangement in which order matters. The number of ways to arrange r objects chosen from n distinct objects is nPr = n! ÷ (n − r)!. To solve a problem, check that order matters, identify n and r, then apply the formula or multiply r descending numbers starting from n.

Understand Permutations of Distinct Objects

A permutation is an arrangement of objects in a definite order. Picking A then B is different from picking B then A. If the order changes, it counts as a new permutation.

Suppose you have 5 different books and want to place 3 of them on a shelf. The first place can be filled in 5 ways. The second place in 4 ways, since one book is used. The third place in 3 ways. Total = 5 × 4 × 3 = 60. This is the multiplication idea behind the formula.

This product is written as nPr, the number of permutations of n distinct objects taken r at a time. It equals n × (n − 1) × (n − 2) × ... for r factors. Using factorials, this is n! ÷ (n − r)!.

When all n objects are arranged (r = n), the answer is n!. For example, 4 different people can stand in a row in 4! = 24 ways. Here the objects must be distinct and repetition is not allowed. Repetition and restrictions are covered in a separate topic.

Key formulas to remember

Permutation of n distinct objects taken r at a time
nPr = n! ÷ (n − r)!
Valid for 0 ≤ r ≤ n, with n and r whole numbers. No repetition.
Product form
nPr = n × (n − 1) × (n − 2) × ... (r factors)
Fastest for calculation. Write exactly r numbers going down from n.
All objects arranged
nPn = n!
Since 0! = 1, this follows from the main formula.
Special values
nP0 = 1, nP1 = n
One way to choose nothing; n ways to fill one place.
Recurrence link
nPr = n × (n−1)P(r−1)
Useful for solving equations such as nPr = k × (n−1)P(r−1).

How to solve Permutations of Distinct Objects questions

Use this method for any question on arranging distinct objects.

  1. 1Read the question and confirm that order or position matters (arrange, rank, seat, form numbers or words, award positions).
  2. 2Check that objects are distinct and that repetition is not allowed.
  3. 3Identify n (total objects available) and r (places to be filled).
  4. 4Write nPr as r factors starting from n and going down by 1 each time.
  5. 5Multiply the factors carefully. Cancel before multiplying if factorials are given.
  6. 6If the question asks for an unknown n or r, form an equation with the product form and solve.
  7. 7Match your answer with the options. As a partial check only: when n and r are known and r ≥ 1, nPr is a multiple of n, so a value that is not a multiple of n is wrong. A multiple of n does not prove your answer is right, so always rely on the calculation.

Quickest way: Descending product method

When to use it: Use for any direct nPr value or any 'arrange r out of n' question where there are no restrictions.

  1. Do not expand full factorials. Write only r numbers: n, n−1, ... down.
  2. For 7P3 write 7 × 6 × 5 = 210.
  3. For all objects, use n! and remember 4! = 24, 5! = 120, 6! = 720, 7! = 5040.
  4. For word problems with all distinct letters, answer is (number of letters)!.
  5. Optional partial check: when n and r are known and r ≥ 1, nPr is a multiple of n. This can catch a slip, but it does not confirm an answer and should not be your main way to choose between options.

Common mistakes in Permutations of Distinct Objects

  • Writing nPr as n! ÷ r! instead of n! ÷ (n − r)!

    Confusion with the combination formula, which has r! in the denominator.

    Fix: Remember that permutation has only (n − r)! below. Check with 5P1, which must be 5.

  • Using more or fewer than r factors in the product

    Students lose count while writing the descending numbers.

    Fix: Count the factors: 8P3 has three numbers, 8 × 7 × 6 = 336.

  • Using permutation when order does not matter

    Words like 'select' or 'choose' are not read carefully.

    Fix: Ask whether swapping two items gives a different result. If yes, use permutation.

  • Treating 0! as 0

    Intuition says factorial of nothing is nothing.

    Fix: Remember 0! = 1. So nPn = n! ÷ 0! = n!.

  • Using the formula for repeated letters in a word

    Students apply n! to words like BALANCE without checking letters are distinct.

    Fix: If any letter repeats, nPr for distinct objects does not apply directly. Check for repeats first.

Worked examples

Example 1

In how many ways can a manager assign 3 different tasks to 3 of 6 available employees, with each employee getting at most one task and the tasks being different?

Show the solution
  1. Tasks are different, so order matters. Assigning employee A to task 1 differs from assigning A to task 2.
  2. n = 6 employees, r = 3 tasks to fill.
  3. Number of ways = 6P3 = 6 × 5 × 4.
  4. 6 × 5 = 30, and 30 × 4 = 120.

Answer: 120 ways

Example 2

How many different 4-letter arrangements can be formed using the letters of the word MONEY?

Show the solution
  1. MONEY has 5 letters, all different: M, O, N, E, Y.
  2. We form 4-letter arrangements, so n = 5 and r = 4.
  3. Number = 5P4 = 5 × 4 × 3 × 2.
  4. 5 × 4 = 20, 20 × 3 = 60, 60 × 2 = 120.

Answer: 120 arrangements

Exam tips

  • Most questions are one-line calculations. Practise 5P2, 6P3, 7P3, 8P2 until you can answer instantly.
  • In letter problems, first scan for repeated letters. If none, the word's letters are all distinct and the formula applies.
  • In equation questions such as nP2 = 56, write n(n − 1) = 56 and solve; then reject any negative or non-integer n.
  • There is no negative marking, so always mark an option. Calculate first; use the multiple-of-n check only as a quick slip detector.
  • Do not spend time expanding large factorials. Cancel common terms first.

Practice questions from Permutation and Combinations

Permutations of Distinct Objects in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Permutations of Distinct Objects: frequently asked questions

What is the formula for nPr in CMA Foundation?

nPr = n! ÷ (n − r)!, where n is the number of distinct objects and r is how many you arrange. In practice, multiply r numbers starting from n and going down by 1.

What is the difference between permutation and combination?

In a permutation, order matters, so AB and BA are different. In a combination, order does not matter, so AB and BA are the same selection. Each combination of r objects gives r! permutations.

How many ways can n distinct objects be arranged in a row?

There are n! ways. This is nPn. For example, 5 distinct objects can be arranged in 5! = 120 ways.

Why is 0! equal to 1?

It is defined as 1 so that formulas stay consistent. For example, nPn = n! ÷ 0! must equal n!, which works only if 0! = 1.