Permutations with Replacement Calculator counts ordered arrangements of r selections from n options when repeats are allowed at every position. The formula is n raised to the power r, since each of the r positions independently has all n options available.
Count ordered selections that allow repeats
Unlike ordinary permutations, which remove an item from the pool once it is chosen, permutations with replacement return every item to the pool after each pick. Choosing a 4-digit PIN where each digit runs from 0 through 9 and digits can repeat is a permutations-with-replacement problem: 10 options for each of 4 positions gives 10^4, equal to 10,000 possible PINs.
Derive the formula from independent choices
Each of the r positions in a with-replacement arrangement has exactly n choices, independent of what was chosen at any other position, because nothing is removed from the pool. Multiplying n by itself r times, once for each independent position, gives n^r as the total count.
Permutations with Replacement Calculator shows this as a single exponentiation rather than a chain of shrinking factors, which is the key difference from ordinary permutations.
Work through a password example
A 3-character password drawn from 26 uppercase letters, with repeats allowed, has 26^3 possible values. Compute 26 × 26 = 676, then 676 × 26 = 17,576. Permutations with Replacement Calculator reports 17,576 directly and shows the exponent form 26^3 alongside it, so the connection between the formula and the multiplication chain stays visible.
Compare with permutations without replacement
Permutations without replacement for the same n and r use n! / (n − r)!, which is always smaller than or equal to n^r, since removing chosen items from the pool reduces the number of options at each later position.
For n = 5, r = 3: permutations without replacement give 5 × 4 × 3 = 60, while permutations with replacement give 5^3 = 125, more than double, because replacement keeps every option available at each of the 3 steps.
Avoid this common mistake
Using the without-replacement formula for a problem that actually allows repeats undercounts the true number of outcomes. A combination lock that allows the same digit to repeat across its positions needs n^r, not n! / (n − r)!.
Read the problem for language like "digits may repeat" or "with replacement" as the signal to use the exponent formula rather than the factorial ratio.
Work through a license-plate example
A license plate format using 3 letters followed by nothing else, drawn from 26 letters with repeats allowed at each position, has 26^3 possible values. Compute 26 × 26 = 676, then 676 × 26 = 17,576, the same structure as the password example above but applied to a different labeling context.
Permutations with Replacement Calculator handles any n and r pair this way, since the underlying math never changes: only the number of independent positions and the size of the pool at each position matter.
Recognize with-replacement language in a word problem
Certain phrases reliably signal a with-replacement problem: "each digit may be used more than once," "selections are made with replacement," "the letter can repeat," or a real-world setup like a die roll, coin flip, or spinner where the same outcome can naturally recur on a later trial.
When none of these signals appear and the problem instead describes selecting people, objects, or items that get "used up," check whether ordinary permutations or combinations are the better fit instead.
Frequently asked questions
What is the formula for permutations with replacement?
The formula for permutations with replacement is n^r, where n is the number of available options and r is the number of ordered positions being filled, with repeats allowed at every position.
How many 4-digit PINs are possible if digits can repeat?
If digits can repeat, a 4-digit PIN drawn from 10 possible digits (0 through 9) has 10^4 = 10,000 possible values.
What is the difference between permutations with and without replacement?
Permutations without replacement remove each chosen item from the pool, using n! / (n − r)!, while permutations with replacement return every item to the pool, using n^r. The with-replacement count is always at least as large.
Can r be greater than n in permutations with replacement?
R can be greater than n in permutations with replacement, since repeats mean there is no upper bound tied to the size of the pool. A 5-character code from a 3-letter alphabet, with repeats, has 3^5 = 243 possible values.
How many 3-letter passwords are possible from 26 letters with repeats allowed?
With repeats allowed, a 3-letter password from 26 letters has 26^3 = 17,576 possible values, computed as 26 × 26 × 26.
Is rolling a die 3 times a permutations-with-replacement problem?
Rolling a die 3 times and recording the ordered sequence of results is a permutations-with-replacement problem, since each roll independently has 6 outcomes regardless of the previous rolls: 6^3 = 216 possible ordered sequences.
How many 3-letter license plate codes are possible from 26 letters with repeats?
With repeats allowed, a 3-letter code drawn from 26 letters has 26^3 = 17,576 possible values, the same calculation used for the password example, since both problems share the identical n = 26, r = 3 structure.
What happens if n or r is entered as zero?
If r is entered as zero, the result is n^0 = 1, since there is exactly one way to fill zero positions: doing nothing. If n is entered as zero while r is positive, the result is 0, since there are no options available to fill any position.
Work through a coin-flip example
Flipping a coin 5 times and recording the ordered sequence of heads and tails is a permutations-with-replacement problem with n = 2 and r = 5: 2^5 = 32 possible ordered sequences, since each flip independently has 2 outcomes regardless of the previous flips.
Summary
Permutations with Replacement Calculator applies n^r to count ordered arrangements where repeats are allowed at every position, treating each of the r positions as an independent choice among n options. Enter n and r to get the exact count, and expect it to exceed the matching without-replacement permutation whenever repeats genuinely expand the set of valid outcomes.