Permutation with Repetition with Calculation Examples
Permutation with repetition counts ordered arrangements when an item may be selected more than once. For r positions filled from n available choices, the count is n^r. A related case occurs when some objects are identical; then the number of distinct arrangements is found by dividing n! by the factorials of the repeated frequencies.
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What Is Permutation with Repetition?
Order matters in a permutation. For example, a 3-digit code made from the digits 1 to 5 has 5 choices for each position, so the number of codes is 5 × 5 × 5 = 5^3 = 125. This formula allows the same digit to appear in more than one position.
Permutation with Repetition Video Lessons
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Permutation with Repetition
Understand how to count arrangements when identical items repeat, including the formula and its application to permutation problems with repeated elements.
Permutation with Repetition Formula & Tricks
Important Formulas
If r ordered positions are filled from n choices and repetition is allowed, multiply n choices for each position.
If r ordered positions are filled from n distinct objects without repetition, the number decreases by one choice after every selection.
If n objects include groups of identical objects with frequencies p1, p2, …, pk, divide n! by the factorial of each repeated frequency.
Quick Tricks
For repetition-allowed arrangements, write the number of choices for each position and multiply them. Equal choices in all r positions give n^r.
Use n^r when an item can be selected again for another position. Use n!/(p1!p2!…) when repeated objects are physically identical and arrangements that swap them are not different.
If one position is fixed, count the remaining positions separately. The fixed position contributes one choice, while each unrestricted position contributes its available choices.
Permutation with Repetition Concepts
Repetition Allowed in Ordered Selections
The multiplication principle gives n choices for position 1, n choices for position 2, and so on, for r positions. The same object may occupy several positions, but changing the order creates a different arrangement.
Arrangements of Identical Objects
If all n objects were treated as distinct, n! arrangements would be counted. Swapping the p identical objects within their group does not create a new arrangement, so division removes these duplicate counts.
Difference Between Repetition Allowed and No Repetition
With repetition allowed, every position has n choices. Without repetition, the choices are n, n − 1, n − 2, and so on. For n = 5 and r = 3, the counts are 5^3 = 125 and 5P3 = 5 × 4 × 3 = 60.
Restricted Permutations with Repetition
A fixed position contributes one choice. If different positions have different numbers of permitted choices, use the product of those numbers rather than n^r. For a code beginning with a fixed digit, only the remaining positions are counted freely.
Permutation with Repetition Data Table
Use the structured data below while following the explanation and examples.
Common Permutation Cases
Formulas and counts for arrangements under different repetition conditions.
| Situation | Formula | Example | Count |
|---|---|---|---|
| r positions from n choices, repetition allowed | n^r | 3 positions from 4 symbols | 4^3 = 64 |
| r distinct objects from n objects, no repetition | n!/(n-r)! | 3 positions from 5 objects | 5P3 = 60 |
| n objects with identical groups | n!/(p1!p2!…pk!) | LEVEL: L twice and E twice | 5!/(2!2!) = 30 |
Note: The first row treats repeated selections as allowed. The third row treats equal objects as indistinguishable.
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Permutation with Repetition: Revision Points
Use the repetition condition and the identity of objects to select the correct counting formula.
- For r positions and n choices with repetition allowed, use n^r.
- Order matters in a permutation, so changing positions changes the arrangement.
- Without repetition, use nPr = n!/(n-r)!.
- For n objects with identical groups of sizes p1, p2, …, pk, use n!/(p1!p2!…pk!).
- For restrictions, multiply the number of permitted choices for each position.
- A PIN may begin with zero because leading zero is allowed in a code unless the question states otherwise.
Permutation with Repetition FAQs
What is the permutation with repetition formula?
If r positions are filled from n choices and repetition is allowed, the formula is n^r.
How many 4-digit PINs can be formed using digits 0 to 9 when repetition is allowed?
Each of the four positions has 10 choices. Therefore, the number of PINs is 10^4 = 10,000.
How is permutation with repetition different from permutation without repetition?
With repetition, the count is n^r. Without repetition, the count is nPr = n!/(n-r)! because the number of available choices falls after each selection.
How many distinct arrangements can be made from the letters of BANANA?
BANANA has 6 letters, with A repeated 3 times and N repeated 2 times. The number of arrangements is 6!/(3!2!) = 60.
What is the number of arrangements of the letters in a word when two letters are identical?
For n total letters with one letter repeated twice, the number of distinct arrangements is n!/2!. For example, AABC gives 4!/2! = 12.
How many 3-letter strings can be made from 5 letters if repetition is allowed?
Each position has 5 choices, so the count is 5^3 = 125.