Choose what to count, then enter the number of kinds n and the number of items r (or a word, for rearranging letters). The formula below is linked to the input fields, so you can also edit the numbers directly in it.
Enter n and r as whole numbers (n from 1 to 1000, r from 0 to 1000). Every digit of the answer is shown exactly, even with dozens of digits. For rearranging letters, you can enter up to 30 characters. Spaces are ignored, and capital and small letters count as different letters.
Result and figure
Choose what to count in the fields on the left, enter the number of kinds and items (or a word), and press "Calculate". The result and a figure will appear here.
Table of Contents
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What you can do on this page
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What is this calculation used for?
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How to Use
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Formulas and figures
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Permutations with repetition \({}_{n}\Pi_{r}\) (arranging \(r\) items, repeats allowed)
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Combinations with repetition \({}_{n}\mathrm{H}_{r}\) (choosing \(r\) items, repeats allowed)
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Distinguishable permutations (rearranging items that include identical ones)
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Which formula should you use? (arrange or just choose × repeats allowed or not)
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Symbols and terms
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Good to know before you start
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How to calculate it in Excel
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How to calculate it in Google Sheets
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How to calculate it in Python
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How to write it in LaTeX and other math languages (copy and paste)
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How to have ChatGPT do the calculation
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DataChef Features
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Related Features
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NumberChef Calculators List
What you can do on this page
- Find the permutations with repetition \({}_{n}\Pi_{r} = n^{r}\), arranging \(r\) items from \(n\) kinds when the same kind can be used any number of times (example - a 4-digit PIN made from the 10 digits 0 to 9)
- Find the combinations with repetition \({}_{n}\mathrm{H}_{r} = {}_{n+r-1}\mathrm{C}_{r}\), choosing \(r\) items when the same kind can be chosen any number of times (example - buying 5 bottles from 3 kinds of juice)
- Just type in a word such as "TOMATO" to count the ways to rearrange letters that include repeats (distinguishable permutations, up to 30 letters)
- Not sure whether to use \(n^{r}\), \({}_{n}\mathrm{H}_{r}\), \({}_{n}\mathrm{P}_{r}\) or \({}_{n}\mathrm{C}_{r}\)? A decision chart based on "Does order matter?" and "Can the same item be chosen again?" is included
- Answers are shown with every digit, no rounding, even with dozens of digits. For long answers, an approximate value such as \(1.2345 \times 10^{18}\) is added
- Diagrams of combinations with repetition using circles and bars (stars and bars), and copy-and-paste formulas for Excel, Google Sheets and Python, are also on this page
If you only need the permutations \({}_{n}\mathrm{P}_{r}\) or combinations \({}_{n}\mathrm{C}_{r}\) without repetition (an item cannot be chosen again once it is picked), see the Permutation and Combination Calculator page.
What is this calculation used for?
Putting a number on how hard a PIN or password is to crack (security)
A 4-digit PIN made from the 10 digits 0 to 9, where the same digit can be used any number of times, has \(10^{4} = 10{,}000\) possibilities. Each extra digit multiplies that by 10, so 6 digits give 1 million.
A password that can use 62 kinds of characters (capital letters, small letters and digits) has \(62^{8} = 218{,}340{,}105{,}584{,}896\) possibilities with 8 characters (about 218 trillion). The permutations with repetition formula lets you compare how much "more kinds of characters" and "a longer password" each help.
How much digital data can express (computers)
Computers represent information with just 2 kinds of symbols, 0 and 1. A byte, a row of 8 of them, can show \(2^{8} = 256\) patterns. That is why a byte is said to have 256 levels.
Full color, with 8 bits each for red, green and blue, gives \(2^{24} = 16{,}777{,}216\) colors, about 16.7 million. The number of screen colors, the bit depth of audio, the number of IP addresses and so on, the "how many can it represent" of every digital device, comes down to permutations with repetition.
Why the genetic code uses 3 letters at a time (life science)
DNA is made of 4 kinds of bases, A, T, G and C, and each group of 3 in a row (a codon) stands for 1 amino acid used to build the body. The same base can repeat, so there are \(4^{3} = 64\) kinds of codons.
The body uses 20 kinds of amino acids, and groups of 2 would give only \(4^{2} = 16\), which is not enough. Permutations with repetition confirm why groups of 3 are enough for life to work.
Counting assortments and orders (retail and food service)
Making a box of a dozen donuts from 5 kinds, with any number of each kind allowed, can be done in \({}_{5}\mathrm{H}_{12} = {}_{16}\mathrm{C}_{12} = 1{,}820\) ways, since order does not matter (combinations with repetition).
Counting orders or inventory breakdowns of "how many of which product" is the same kind of calculation. You can estimate with real numbers, not just a feeling, how many more patterns you need to manage when you add one more product.
Counting shortest paths (logistics and circuit design)
On a grid of city blocks, the number of shortest paths that go 3 blocks right and 2 blocks up is the same as the number of rearrangements of the 5 letters "R, R, R, U, U", so it is \(\dfrac{5!}{3!\,2!} = 10\). Distinguishable permutations work directly here.
This calculation is the basis for estimating the number of possible delivery routes or wiring patterns on a circuit board, and it shows in numbers how explosively the options grow as the grid gets larger.
Formulas and figures
Permutations with repetition \({}_{n}\Pi_{r}\) (arranging \(r\) items, repeats allowed)
Figure
Standard notation
(the usual math form)
\({}_{n}\Pi_{r}\)
\(=\)
\(n\)
\(r\)
In words
(symbols replaced with words)
③
\({}_{n}\Pi_{r}\): ways to arrange \(r\) items with repeats allowed
\(=\)
①
\(n\): number of kinds for each place
②
\(r\): number of items to arrange
The formula in words
①
Multiply the
\(n\): number of kinds for each place
by itself
②
as many times as the
\(r\): number of items to arrange
③
and you get the
\({}_{n}\Pi_{r}\): ways to arrange \(r\) items with repeats allowed
Quick example
The number of ways to make a 4-digit PIN from the 10 digits 0 to 9 is
ways to make a 4-digit PIN
\(=\)
digits for each place (10)
number of digits (4)
\({}_{10}\Pi_{4} = 10^{4} = 10 \times 10 \times 10 \times 10 = 10000\)
Key idea
Since the same item can be chosen any number of times, the number of choices stays at \(n\) for the second place, the third place and so on. It never goes down. So you multiply by \(n\) every time, and multiplying \(n\) by itself \(r\) times gives \(n^{r}\).
This is the only difference from the permutations \({}_{n}\mathrm{P}_{r} = n \times (n-1) \times \cdots\), where the choices go down by 1 each time you pick. Ask yourself "Does the supply run out or not?" and you can tell them apart.
The \(\Pi\) in \({}_{n}\Pi_{r}\) is the capital Greek letter pi, which stands for a product (multiplication). It matches the P in "product".
However, this symbol is not used nearly as widely as \({}_{n}\mathrm{P}_{r}\) and \({}_{n}\mathrm{C}_{r}\). Most US textbooks simply say that the number of permutations with repetition is \(n^{r}\). On homework and tests, writing \(n^{r}\) is always understood.
Combinations with repetition \({}_{n}\mathrm{H}_{r}\) (choosing \(r\) items, repeats allowed)
Figure
Standard notation
(the usual math form)
\({}_{n}\mathrm{H}_{r}\)
\(=\)
\(n+r-1\)
\(\mathrm{C}\)
\(r\)
In words
(symbols replaced with words)
③
ways to choose \(r\) items with repeats allowed
\(=\)
①
\(n+r-1\): places for circles and bars
\(\mathrm{C}\)
②
\(r\): places that get a circle
The formula in words
①
Line up \(r\) circles and \(n-1\) bars that mark the boundaries between kinds. Out of the
\(n+r-1\): places for circles and bars in all
,
②
count the ways to pick the
\(r\): places that get a circle
③
and you get the
\({}_{n}\mathrm{H}_{r}\): ways to choose \(r\) items with repeats allowed
Quick example
The number of ways to buy 5 bottles from 3 kinds of juice (apple, orange and grape), with any number of each kind allowed, is
ways to buy 5 from 3 kinds
\(=\)
places (3 + 5 − 1 = 7)
\(\mathrm{C}\)
places for circles (5)
\({}_{3}\mathrm{H}_{5} = {}_{3+5-1}\mathrm{C}_{5} = {}_{7}\mathrm{C}_{5} = \dfrac{7 \times 6}{2 \times 1} = 21\)
Key idea
We were only choosing, so why does it turn into the combination \({}_{n+r-1}\mathrm{C}_{r}\)? The secret is rewriting the choice as circles and bars (a method known as "stars and bars"). Before memorizing it as a formula, follow the steps below.
First, fix the order of the kinds: apple, orange, grape. Then write each bottle you buy as a circle ○ and each boundary between kinds as a bar |. Buying 2 apple, 1 orange and 2 grape becomes ○○|○|○○. Buying 3 apple, 0 orange and 2 grape becomes ○○○||○○. A kind you buy none of just shows up as two bars next to each other, so it can be written too.
Going the other way, if you are shown a row of circles and bars first, you can read exactly one way of buying from it. Look at the 3 spaces separated by the bars, from left to right. The number of circles in each space is the number of apple, orange and grape bottles. ○○||○○○ means 2 apple, 0 orange and 3 grape. From a purchase to a row of symbols, and from a row of symbols back to a purchase, there is always exactly one answer. So purchases and rows of symbols match up one to one, with nothing missed and nothing counted twice.
Now all you need to do is count the rows of symbols. The bars separate 3 kinds, so there are 2 of them (\(n-1\) in general), and there are \(r\) circles, one for each bottle. That makes \(n+r-1\) symbols in all, in places numbered 1, 2, … from the left. Choosing one row is the same as choosing which \(r\) of the \(n+r-1\) numbered places get a circle. At this point it is just an ordinary combination, where the same place cannot be chosen twice. That is why \({}_{n}\mathrm{H}_{r} = {}_{n+r-1}\mathrm{C}_{r}\).
You can also choose the places for the bars instead of the circles. The remaining places automatically get circles, so it comes to the same thing. Writing \({}_{n+r-1}\mathrm{C}_{n-1}\) gives the same answer.
The \(\mathrm{H}\) is said to come from "homogeneous", as in the collection of all terms of the same degree. The symbol \({}_{n}\mathrm{H}_{r}\) is rarely used in US textbooks. There, you simply write the answer as \({}_{n+r-1}\mathrm{C}_{r}\) or \(\binom{n+r-1}{r}\), using stars and bars as above.
Distinguishable permutations (rearranging items that include identical ones)
Figure
Standard notation
(the usual math form)
\(N\)
\(=\)
\(n!\)
\(\div\)
\(p!\,q!\,r!\cdots\)
In words
(symbols replaced with words)
③
\(N\): ways to arrange \(n\) items that include identical ones
\(=\)
①
\(n!\): arrangements if every item were different
\(\div\)
②
\(p! \times q! \times \cdots\): ways to reorder identical items among themselves
The formula in words
①
Take the
\(n!\): arrangements if every item were different
②
divide it by the
\(p! \times q! \times \cdots\): ways to reorder identical items among themselves
to remove the repeats
③
and you get the
\(N\): ways to arrange \(n\) items that include identical ones
Quick example
The number of strings you can make by rearranging the 6 letters of "TOMATO" (two Ts, two Os, one M and one A) is
rearrangements of TOMATO
\(=\)
6 letters all different (6!)
\(\div\)
reorderings of identical letters (2! × 2! × 1! × 1!)
\(N = \dfrac{6!}{2!\,2!\,1!\,1!} = \dfrac{720}{2 \times 2 \times 1 \times 1} = \dfrac{720}{4} = 180\)
Key idea
Give the two Ts temporary labels, \(\mathrm{T}_1\) and \(\mathrm{T}_2\), and all 6 letters become different, so there are \(6! = 720\) arrangements. But once you erase the labels, \(\mathrm{T}_1\mathrm{T}_2\) and \(\mathrm{T}_2\mathrm{T}_1\) look exactly the same, so each arrangement has been counted twice.
If a letter appears \(p\) times, all \(p!\) ways to reorder those \(p\) copies look the same, so you divide by \(p!\) for each kind of letter. A letter that appears only once gives \(1! = 1\), so writing it or not does not change the answer.
Be careful about what the letters mean here. Unlike the \(n\) in the two formulas above (the number of kinds you choose from), the \(n\) in this formula is the total number of items to arrange. "TOMATO" has 4 kinds of letters, T, O, M and A, but you arrange 6 letters, so \(n = 6\).
The \(p,\ q,\ r,\ \dots\) in the denominator are the counts for each kind, so they always add up to \(p + q + r + \cdots = n\). For TOMATO, \(2 + 2 + 1 + 1 = 6\), which matches the number of letters. If this sum does not match, it is a sign that you missed something or counted something twice. (The \(r\) here is just the third letter used for a count. It is not the "number of items \(r\)" from the formulas above.)
\(\dfrac{n!}{p!\,q!\cdots}\) is also called a multinomial coefficient. Along with combinations with repetition, it is a classic way to count arrangements that include identical items. The number of shortest paths on a grid of city blocks that go 3 blocks right and 2 blocks up, \(\dfrac{5!}{3!\,2!} = 10\), is another use of this formula.
Which formula should you use? (arrange or just choose × repeats allowed or not)
Figure
Standard notation
(the usual math form)
\({}_{n}\Pi_{r} = n^{r}\)
\({}_{n}\mathrm{H}_{r} = {}_{n+r-1}\mathrm{C}_{r}\)
\({}_{n}\mathrm{P}_{r} = \dfrac{n!}{(n-r)!}\)
\({}_{n}\mathrm{C}_{r} = \dfrac{n!}{r!\,(n-r)!}\)
In words
(symbols replaced with words)
①
arrange × the same item can be chosen any number of times
②
just choose × the same item can be chosen any number of times
③
arrange × an item cannot be chosen again once it is picked
④
just choose × an item cannot be chosen again once it is picked
The formula in words
①
If order matters (role, rank or digit place) and the same item can be chosen any number of times, use the
permutations with repetition \({}_{n}\Pi_{r} = n^{r}\)
②
If only the selection matters and the same item can be chosen any number of times, use the
combinations with repetition \({}_{n}\mathrm{H}_{r} = {}_{n+r-1}\mathrm{C}_{r}\)
③
If order matters and an item cannot be chosen again once it is picked, use the
permutations \({}_{n}\mathrm{P}_{r}\)
④
If only the selection matters and an item cannot be chosen again once it is picked, use the
combinations \({}_{n}\mathrm{C}_{r}\)
Quick example
Even with the same numbers 4 and 3, the formula changes with the situation
ways to buy 3 scoops from 4 ice cream flavors (any flavor can repeat)
\(=\)
combinations with repetition
ways to pick 1st, 2nd and 3rd place from 4 people
\(=\)
permutations
\({}_{4}\mathrm{H}_{3} = {}_{6}\mathrm{C}_{3} = 20\)
\({}_{4}\mathrm{P}_{3} = 4 \times 3 \times 2 = 24\)
Key idea
Before you get to these 4, there is one fork in the road. Is it a situation where you choose \(r\) items from \(n\) kinds? Or is it one where the items are already in front of you and you arrange all of them, like the 6 letters of TOMATO? In the second case there is no choosing, so instead of the 4 formulas above, use distinguishable permutations \(\dfrac{n!}{p!\,q!\cdots}\). First, tell apart "choose, then arrange" and "arrange everything you have".
Once you know it is a choosing situation, check these 2 questions in this order. (1) Are you arranging, or only counting the selection? (2) Can the same item be chosen two or more times, or is it done once it is picked?
The clue for (1) is whether each place has its own meaning. If the places have roles, such as 1st and 2nd place, captain and vice captain, or the 1st and 2nd digits of a PIN, you are arranging. If only the group matters, such as 2 people on cleanup duty or what is in a shopping basket, you are just choosing.
The clue for (2) is whether the supply runs out. If choosing does not use anything up, as when you roll a die again and again, buy any number of the same flavor, or reuse the same digit, repeats are allowed. If it does, as when one person cannot hold 2 roles or the same ticket cannot be drawn twice, there are no repeats.
The gap between \(n^{r}\) and \({}_{n}\mathrm{P}_{r}\) widens as \(r\) grows (choosing 4 from 10 kinds gives \(10^{4} = 10000\) versus \({}_{10}\mathrm{P}_{4} = 5040\)). Mixing up (2) can nearly double the answer, so do not skip this check.
To arrange with repeats allowed, use the permutations with repetition \({}_{n}\Pi_{r} = n^{r}\). To just choose with repeats allowed, use the combinations with repetition \({}_{n}\mathrm{H}_{r} = {}_{n+r-1}\mathrm{C}_{r}\). To arrange everything you have when some items are identical, use \(\dfrac{n!}{p!\,q!\cdots}\). The key idea of this topic is that rewriting a combination with repetition as circles and bars (stars and bars) turns it into an ordinary combination \({}_{n+r-1}\mathrm{C}_{r}\).
Symbols and terms
Symbols
| \(n\) | en | The number of kinds you choose from. (Example - choosing from 3 kinds of juice, \(n = 3\); using the digits 0 to 9, \(n = 10\).) The letter \(n\), for "number", is often used for counts. Only in distinguishable permutations \(\dfrac{n!}{p!\,q!\cdots}\) does it mean something else: the total number of items to arrange ("TOMATO" has 4 kinds of letters, but \(n = 6\)). |
| \(r\) | ar | The number of items you choose (or arrange) with repeats allowed. (Example - buying 5 bottles of juice, \(r = 5\); a 4-digit PIN, \(r = 4\).) It sits in the same position as in the permutation and combination symbols \({}_{n}\mathrm{P}_{r}\) and \({}_{n}\mathrm{C}_{r}\). |
| \(n!\) | n factorial | The product of all the whole numbers from \(n\) down to \(1\). It is the number of ways to arrange all \(n\) items in a row. By definition, \(0! = 1\). (Example - \(4! = 4 \times 3 \times 2 \times 1 = 24\)) |
| \({}_{n}\Pi_{r}\) | n pi r | Permutations with repetition: the number of ways to choose \(r\) items from \(n\) kinds with repeats allowed and arrange them in order. Its value is \(n^{r}\). \(\Pi\) is the capital Greek letter pi, which stands for a product and matches the P in "product". Most US textbooks just write \(n^{r}\) without this symbol, so follow the notation used in your class. |
| \({}_{n}\mathrm{H}_{r}\) | n H r | Combinations with repetition: the number of ways to choose \(r\) items from \(n\) kinds with repeats allowed, where order does not matter. Its value is \({}_{n+r-1}\mathrm{C}_{r}\). The \(\mathrm{H}\) is said to come from "homogeneous", as in the collection of all terms of the same degree. This symbol is rarely seen in US textbooks, which write \(\binom{n+r-1}{r}\) instead. |
| \({}_{n}\mathrm{P}_{r}\) | n P r | Permutations: the number of ways to choose \(r\) of \(n\) items and put them in order, where an item cannot be chosen again once it is picked. P stands for "permutation". |
| \({}_{n}\mathrm{C}_{r}\) | n C r | Combinations: the number of ways to choose \(r\) of \(n\) items, where an item cannot be chosen again once it is picked and order does not matter. C stands for "combination". |
| \(N\) | capital N | On this page, the letter for the number of distinguishable permutations. It is the capital first letter of "number" and is often used for a total you want to find. |
| \(p,\ q\) | p, q | In distinguishable permutations, the letters for how many copies of each identical item there are. (Example - in "TOMATO", T appears \(p = 2\) times, O \(q = 2\) times, M once and A once.) With 3 or more kinds, continue with \(p,\ q,\ r,\ \dots\) (this \(r\) is not the "number of items \(r\)"). They always add up to \(p + q + r + \cdots = n\), the total number of items to arrange. |
| \(\bigcirc\) and \(|\) | circle and bar | The symbols used to count combinations with repetition (a method known as "stars and bars"). A circle ○ is one item chosen, and a bar \(|\) is a boundary between kinds. ○○|○|○○ means 2 of the first kind, 1 of the second and 2 of the third. |
Terms
| counting | Finding how many possible outcomes there are in total. Permutations, combinations and their versions with repetition are all tools for counting efficiently. |
| repetition | The same item appearing two or more times. In this topic, it means that the same item may be chosen two or more times. It is also described as choosing "with replacement". |
| permutation with repetition | An arrangement of \(r\) items chosen from \(n\) kinds, where the same kind can be chosen any number of times and order matters. There are \(n^{r}\) of them, written \({}_{n}\Pi_{r}\) as a symbol. |
| combination with repetition | A selection of \(r\) items from \(n\) kinds, where the same kind can be chosen any number of times and order does not matter. There are \({}_{n}\mathrm{H}_{r} = {}_{n+r-1}\mathrm{C}_{r}\) of them. |
| distinguishable permutation | An arrangement of all \(n\) items when some of them are identical, as in "TOMATO". There are \(\dfrac{n!}{p!\,q!\cdots}\) of them: you divide by the reorderings of identical items to remove the repeats. Also called a permutation with identical items. |
| permutation | A way to choose \(r\) of \(n\) items and put them in a row, where order matters. The symbol is \({}_{n}\mathrm{P}_{r}\), and an item cannot be chosen again once it is picked. |
| combination | A way to choose \(r\) of \(n\) items where order does not matter. The symbol is \({}_{n}\mathrm{C}_{r}\), and an item cannot be chosen again once it is picked. |
| factorial | Multiplying all the whole numbers from \(n\) down to \(1\). The symbol is \(n!\), and it gives the number of ways to arrange all \(n\) items. |
| stars and bars | The method of counting combinations with repetition with symbols: a bar \(|\) marks each boundary between kinds, and a star (a circle ○ on this page) marks each item. Separating \(n\) kinds takes \(n-1\) bars, so the problem becomes choosing the places for the circles out of \(n+r-1\) places. |
| multinomial coefficient | Another name for the number of distinguishable permutations, \(\dfrac{n!}{p!\,q!\cdots}\). The name comes from the fact that the coefficients you get when you expand \((a+b+c)^{n}\) have this form. |
| tree diagram | A branching diagram that lists every possible outcome, so you count each one exactly once. Permutations with repetition correspond to a tree where every branch splits into the same number of branches. |
| shortest path | On a grid of streets, a route to the destination with no detours. The number of shortest paths that go \(p\) blocks right and \(q\) blocks up is the distinguishable permutations \(\dfrac{(p+q)!}{p!\,q!}\). |
Good to know before you start
Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back over the topics in this table is the fastest way forward.
| Multiplication and division (Grades 3–5) |
|
| Powers and exponents (Grades 6–8, Algebra 1) |
|
| Counting outcomes and tree diagrams (Grade 7) |
|
| Factorials (high school, Algebra 2 or Precalculus) |
|
| Permutations and combinations (high school, Algebra 2 or Statistics) |
|
How to calculate it in Excel
Copy the whole table below and paste it into cell A1 in Excel. It works as is.
Table to find the permutations with repetition nΠr
| Number of kinds n | 10 |
| Items to arrange r | 4 |
| Permutations with repetition nΠr | =B1^B2 |
Table to find the combinations with repetition nHr
| Number of kinds n | 3 |
| Items to choose r | 5 |
| Places for circles and bars | =B1+B2-1 |
| Combinations with repetition nHr | =COMBIN(B3,B2) |
Table to find the distinguishable permutations
| Total letters n (TOMATO) | 6 |
| Number of Ts | 2 |
| Number of Os | 2 |
| Number of Ms | 1 |
| Number of As | 1 |
| Arrangements N | =FACT(B1)/(FACT(B2)*FACT(B3)*FACT(B4)*FACT(B5)) |
After pasting, the upper rows are your inputs and the last row is calculated automatically.
"^" is the power symbol, FACT is the factorial, and COMBIN is the Excel function for combinations (nCr).
The first table is the 4-digit PIN from the 10 digits 0 to 9, and B3 shows 10000.
The second table is buying 5 bottles from 3 kinds of juice. B3 shows 7 (= 3 + 5 − 1) and B4 shows 21.
The third table rearranges "TOMATO", and B6 shows 180. If there are more kinds of repeated letters, multiply in more FACT terms in the denominator. =MULTINOMIAL(B2,B3,B4,B5) also gives the same 180.
Note that Excel keeps only 15 significant digits, so when the answer has 16 or more digits, the last digits are rounded.
"^" is the power symbol, FACT is the factorial, and COMBIN is the Excel function for combinations (nCr).
The first table is the 4-digit PIN from the 10 digits 0 to 9, and B3 shows 10000.
The second table is buying 5 bottles from 3 kinds of juice. B3 shows 7 (= 3 + 5 − 1) and B4 shows 21.
The third table rearranges "TOMATO", and B6 shows 180. If there are more kinds of repeated letters, multiply in more FACT terms in the denominator. =MULTINOMIAL(B2,B3,B4,B5) also gives the same 180.
Note that Excel keeps only 15 significant digits, so when the answer has 16 or more digits, the last digits are rounded.
How to calculate it in Google Sheets
Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.
Table to find the permutations with repetition nΠr
| Number of kinds n | 10 |
| Items to arrange r | 4 |
| Permutations with repetition nΠr | =B1^B2 |
Table to find the combinations with repetition nHr
| Number of kinds n | 3 |
| Items to choose r | 5 |
| Places for circles and bars | =B1+B2-1 |
| Combinations with repetition nHr | =COMBIN(B3,B2) |
Table to find the distinguishable permutations
| Total letters n (TOMATO) | 6 |
| Number of Ts | 2 |
| Number of Os | 2 |
| Number of Ms | 1 |
| Number of As | 1 |
| Arrangements N | =FACT(B1)/(FACT(B2)*FACT(B3)*FACT(B4)*FACT(B5)) |
The same formulas as in Excel work as is (FACT, COMBIN and MULTINOMIAL have the same names in Google Sheets).
Copy the whole table, paste it into cell A1, and change the numbers to fit your situation.
Copy the whole table, paste it into cell A1, and change the numbers to fit your situation.
How to calculate it in Python
import math
from collections import Counter
# 1) permutations with repetition (choose r from n kinds with repeats allowed, and order matters)
kinds = 10 # number of kinds n (the 10 digits 0-9)
picks = 4 # number of items to arrange r (4 digits)
repeated_permutations = kinds ** picks
# 2) combinations with repetition (choose r from n kinds with repeats allowed, order does not matter)
juice_kinds = 3 # number of kinds n (3 kinds of juice)
bottles = 5 # number of items to choose r (5 bottles)
repeated_combinations = math.comb(juice_kinds + bottles - 1, bottles)
# 3) distinguishable permutations (rearranging a word with repeated letters)
word = "TOMATO"
letter_counts = Counter(word)
same_item_permutations = math.factorial(len(word))
for count in letter_counts.values():
same_item_permutations //= math.factorial(count)
print(f"Permutations with repetition (arrange {picks} from {kinds} kinds): {repeated_permutations}")
print(f"Combinations with repetition (choose {bottles} from {juice_kinds} kinds): {repeated_combinations}")
print(f"Rearrangements of {word}: {same_item_permutations}")
Runs with the standard library only (math.comb needs Python 3.8 or later). Running it shows 10000, 21 and 180. Python integers have no limit on the number of digits, so the answer stays exact even with hundreds of digits. Change the numbers and the word at the top and run it.
How to write it in LaTeX and other math languages (copy and paste)
Permutations with repetition \({}_{n}\Pi_{r}\) (arranging \(r\) items, repeats allowed)
ₙΠᵣ = nʳ
{}_{n}\Pi_{r} = n^{r}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mmultiscripts>
<mi>Π</mi>
<mi>r</mi><none/>
<mprescripts/>
<mi>n</mi><none/>
</mmultiscripts>
<mo>=</mo>
<msup><mi>n</mi><mi>r</mi></msup>
</mrow>
</math>
n^r
n^r
repeatedPermutation := n^r;
repeated_permutation = n^r;
n^r
Combinations with repetition \({}_{n}\mathrm{H}_{r}\) (choosing \(r\) items, repeats allowed)
ₙHᵣ = ₙ₊ᵣ₋₁Cᵣ
{}_{n}\mathrm{H}_{r} = {}_{n+r-1}\mathrm{C}_{r} = \dfrac{(n+r-1)!}{r!\,(n-1)!}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mmultiscripts>
<mi>H</mi>
<mi>r</mi><none/>
<mprescripts/>
<mi>n</mi><none/>
</mmultiscripts>
<mo>=</mo>
<mmultiscripts>
<mi>C</mi>
<mi>r</mi><none/>
<mprescripts/>
<mrow><mi>n</mi><mo>+</mo><mi>r</mi><mo>−</mo><mn>1</mn></mrow><none/>
</mmultiscripts>
</mrow>
</math>
H(n, r) = C(n+r-1, r)
Binomial[n + r - 1, r]
nHr := binomial(n + r - 1, r);
nhr = nchoosek(n + r - 1, r);
H(n,r) = C(n+r−1, r) = (n+r−1)!/(r!(n−1)!)
Distinguishable permutations (rearranging items that include identical ones)
N = n! ÷ (p! × q! × r! × ⋯)
N = \dfrac{n!}{p!\,q!\,r!\cdots}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mi>N</mi>
<mo>=</mo>
<mfrac>
<mrow><mi>n</mi><mo>!</mo></mrow>
<mrow>
<mi>p</mi><mo>!</mo>
<mi>q</mi><mo>!</mo>
<mi>r</mi><mo>!</mo>
<mo>⋯</mo>
</mrow>
</mfrac>
</mrow>
</math>
N = (n!)/(p!q!r!cdots)
Multinomial[p, q, r]
N := factorial(n)/(factorial(p)*factorial(q)*factorial(r));
N = factorial(n)/(factorial(p)*factorial(q)*factorial(r));
N = n!/(p!q!r!⋯)
How to have ChatGPT do the calculation
You are a calculation assistant for counting (permutations and combinations). Do the following 3 calculations by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing). 1. How many 4-digit PINs can be made from the 10 digits 0 to 9? (The same digit can be used any number of times.) 2. How many ways are there to buy 5 bottles from 3 kinds of juice, with any number of each kind allowed? (Count only how many of each kind, not the order of choosing.) 3. How many different strings can be made by rearranging the 6 letters of "TOMATO"? For each, say whether you used permutations with repetition, combinations with repetition or distinguishable permutations, explain why, and show the numbers from the execution result. In Python, use the math module (math.comb and math.factorial) and collections.Counter.
How to Use
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1Enter your numbersType the numbers you want to calculate with into the input fields
-
2CalculatePress the "Calculate" button
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3Check the resultThe result appears on the spot. The same page also explains the idea behind the calculation and the formula
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