Permutation Calculation Practice Problems and Solutions

Permutation calculation practice problems help you count arrangements when order matters. Enter a total number of distinct items, n, and the number chosen, r; this calculator returns the number of ordered arrangements, nPr, and the corresponding combination count, nCr, for comparison.

Formula

The number of ways to select and arrange r distinct items from n distinct items is the permutation formula:

nPr = n! / (n − r)!

Here, n is the total number of available distinct items, r is the number placed in the arrangement, and the exclamation mark means factorial: n! = n × (n − 1) × … × 2 × 1, with 0! = 1. Both inputs are counts, so they have no physical unit. The result is a count of arrangements.

The comparison output uses the combination formula, which counts selections without regard to order:

nCr = n! / [r!(n − r)!]

For valid inputs, n and r are nonnegative whole numbers and r ≤ n. Equivalently, permutations can be calculated as a product of r descending factors: n × (n − 1) × … × (n − r + 1). This form makes clear why choosing a first position, then a second, generally gives different arrangements.

Worked Example

Using the calculator defaults, n = 5 and r = 3. Suppose five distinct books are available and three will be arranged on a shelf. There are five choices for the first position, four remaining choices for the second, and three for the third:

5P3 = 5 × 4 × 3 = 60 arrangements.

The factorial form gives the same result: 5! / (5 − 3)! = 120 / 2 = 60. If order is ignored, the number of groups of three is 5! / [3! × 2!] = 10. Thus the calculator shows 60 permutations and 10 combinations. Each combination corresponds to 3! = 6 possible arrangements, and 10 × 6 = 60.

Practice Problems

Try each question before reading its answer. Unless a problem says otherwise, assume every item is distinct and that no item can be used twice.

1. Four finalists: In how many ways can first, second, and third place be assigned among four finalists? Answer: 4P3 = 4 × 3 × 2 = 24.
2. Three letters: How many three-letter arrangements can be made from A, B, C, D, and E without repeating a letter? Answer: 5P3 = 5 × 4 × 3 = 60.
3. All six people: How many ways can six distinct people stand in a line? Answer: 6P6 = 6! = 720.
4. Choose a committee: How many groups of three can be selected from eight people if their roles do not matter? This is not a permutation: 8C3 = 8! / (3!5!) = 56 groups.
5. Assign two roles: A club has seven members. How many ways can it assign a president and a secretary to different members? The roles differ, so order matters: 7P2 = 7 × 6 = 42.

How to Choose the Right Counting Method

The central question is whether changing the order creates a new outcome. If Ana, Ben, and Chen are assigned gold, silver, and bronze places, swapping two people changes the result. Use a permutation. If those same three people form an unranked committee, swapping their names does not create a different group. Use a combination.

• Use a permutation for ordered positions, rankings, schedules, or distinct roles.
• Use a combination for unordered selections such as a committee or a set of chosen ingredients.
• Use a product rule when a problem has stages with different choices at each stage. For example, three distinct positions filled from five items give 5 × 4 × 3.

A useful check is to ask whether two outcomes that contain the same selected items but list them in different orders should count separately. If yes, order matters. If no, it does not.

Repeated Items and Other Variations

The standard nPr formula assumes all available items are distinct and each item is used at most once. If repetition is allowed and each of r positions can be filled by any of n choices, there are n^r sequences. For example, a four-character code made from ten digits, with digits allowed to repeat, has 10⁴ = 10,000 possible codes. That is not the same problem as arranging four distinct digits.

If all items are arranged but some are identical, the basic n! count overcounts arrangements that look the same. For a collection of n items with repeated groups of sizes a, b, and so on, the number of distinct arrangements is n! / (a!b!…). For example, the letters in LEVEL have five positions, with L repeated twice and E repeated twice, giving 5!/(2!2!) = 30 distinct arrangements.

Common Mistakes and Limits

• Ignoring order when it matters: A ranking or assignment of different roles is ordered even if the same people appear in every outcome.
• Counting order when it does not matter: Divide nPr by r! to obtain nCr when every ordering of a chosen group represents the same selection.
• Using invalid inputs: The usual formula requires whole-number counts with 0 ≤ r ≤ n. A request to arrange more distinct items than are available is not possible without repetition.
• Forgetting the assumptions: The calculator treats items as distinct and disallows reuse. Identical objects or repetition require a different counting model.

Factorials grow quickly, so large answers can become difficult to calculate by hand. For modest problems, multiplying only the r descending factors is often simpler than calculating two large factorials. In a real counting problem, also define exactly what counts as a different outcome before applying a formula; the arithmetic cannot resolve an unclear setup.

Frequently Asked Questions

What is the difference between a permutation and a combination?
A permutation counts ordered arrangements; a combination counts selections without regard to order. For the same n and r, nPr = nCr × r!.

How do you calculate nPr by hand?
Multiply n by the next r − 1 descending whole numbers. For example, 8P3 = 8 × 7 × 6 = 336.

Is 0! equal to 1?
Yes. This standard definition makes factorial formulas work consistently, including cases such as selecting and arranging zero items: 0P0 = 1.

Can a permutation have repeated items?
The basic nPr formula assumes no repeats. If choices may be reused independently in r positions, use n^r; if objects are identical, account for the repeated groups separately.

When should I divide a permutation by r!?
Divide by r! when the problem first counts every ordering of a selected group but the final result should treat those orders as the same selection. This converts nPr into nCr.

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in n | Total distinct items | - | 5 | 0
in r | Items arranged | - | 3 | 0
out arrangements | Permutations (nPr) | arrangements | npr(n,r) | 0
out combinations | Combinations (nCr) | groups | ncr(n,r) | 0