Permutation with repetition means arranging items where some or all of them can repeat. In simpler terms, it's about figuring out how many different ways you can arrange a group of things when you’re allowed to use the same item more than once.

For example, if you have 3 digits (1, 2, and 3) and want to make 2-digit numbers using them, you can have numbers like 11, 12, 13, 21, 22, etc. Since repetition is allowed, the same digit can be used again in the arrangement.

In this concept, we use a formula to find the number of possible arrangements. The general formula is:
n^r, where n is the number of items to choose from, and r is the number of positions to fill.

This type of permutation is commonly used in passwords, PINs, and codes where characters can be repeated. We’ll also explore circular permutations, formulas, and examples to understand it better.

What is Permutation?

Permutation means arranging a group of items in a specific order. These items can be anything like numbers, letters, colors, or symbols. For example, if you are choosing who comes first, second, and third in a race, that is a permutation because the order matters.

In math, permutation refers to placing the elements of a set into a particular sequence. The total number of ways you can do this depends on how many items there are and whether some of them repeat.

Permutations are used in many areas of mathematics and real-life situations, like forming different codes, arranging books on a shelf, or deciding seating plans. When solving such problems, we use a specific formula called the permutation formula.

There are also some special properties and types of permutations that help us understand and solve different arrangement problems more easily.

Learn the various concepts of the Binomial Theorem here.

Permutation with Repetition

The Permutation with Repetition is the simplest to determine. Consider when a piece has n different types and one has r choices each time then the permutations is defined by: n × n × … (r times). This implies there are n possibilities for the first selection, followed by n possibilities for the second selection, and so on, multiplying each time.

Permutation with Repetition Formula

The number of permutations of n objects, where p objects are of one kind, q objects are of another kind and the rest, if any, are of a different kind is nPr / (p! × q!)

Check out this article on Arithmetic Mean.

How do you solve permutation with repetition?

Consider the word TEETH. There are 2 E's in the word. Both E's are identical, and it does not matter in which order we write these 2 E's since they are the same. In other words, if we exchange ‘E’ for ‘E’, we still spell TEETH. The same is true for the T's since there are 2 T's in the word TEETH as well. In how many ways can we arrange the letters in the word TEETH?

We must account for the fact that these 2 E's are identical and that the 2 T's are identical. We do this using the formula:

nPr / (x1! × x2!), where x is the number of times a letter is repeated.

Using the word TEETH:
n = 5 (letters), repeated letters: E appears 2 times, T appears 2 times

So:
5P5 / (2! × 2!) = 5 × 4 × 3 × 2 × 1 / (2 × 1 × 2 × 1) = 120 / 4 = 30

Circular permutations with repetition

The formula for circular permutations with repetition for n elements is:

n! / n = (n−1)!

Let us determine the number of distinguishable permutations of the letters ELEMENT.

Suppose we make all the letters different by labelling the letters as follows:

E1, L, E2, M, E3, N, T

Now, all the letters are different from each other. In this case, there are (7 - 1)! = 6! = 720 different circular permutations possible.

Let us consider one such circular permutation where the elements are arranged in the order:

L, E1, M, E2, N, E3, T

We can form new arrangements of permutations from this arrangement by only moving the E's. Clearly, there are 3! = 6 such arrangements. We list them below:

LE1ME2NE3T
LE1ME3NE2T
LE2ME1NE3T
LE2ME3NE1T
LE3ME2NE1T
LE3ME1NE2T

But since the E's are identical, all these 6 arrangements count as just one: LEMENET

This is true for every permutation involving identical letters.

1. Order Matters
   In permutations, the order in which elements are arranged is important. For example, "ABC" is different from "CAB".

2. Repetition is Allowed
   You can use the same element more than once in the arrangement. For example, in 3-digit numbers using digits 1 to 5, "111" is valid.

3. Formula Used
   If you have n choices and you want to choose r elements, then the number of permutations with repetition is:
   nr

4. Total Number of Arrangements
   Each position in the arrangement can be filled in n ways, and since repetition is allowed, all positions can use the same or different values.

5. Applicable in Real-Life Scenarios
   Permutations with repetition apply in situations like:

   • Creating passwords (letters/digits can repeat)

   • Generating color codes

   • Assigning lockers with repeated numbers

6. Used in Solving Counting Problems
   These permutations help solve problems involving sequences, combinations with repetition, and digital or character-based arrangements.

Solved Examples on Permutation with Repetition

Example 1: Find the number of permutations of the letters of the word MICROSOFT.

Solution:

The word MICROSOFT consists of 9 letters, in which the letter ‘O’ is repeated two times. Therefore, the number of permutations of the letters of the word MICROSOFT = 9! / 2! = 181440.

Example 2: How many arrangements can be made, with the letters of the word CALCULATOR? In how many of these arrangements, vowels occur together?

Solution:

The word CALCULATOR consists of 10 letters, in which ‘C’ is repeated two times, ‘A’ is repeated two times, ‘L’ is repeated two times and the rest all are different. Therefore, the number of permutations of the letters of the word CALCULATOR = 10! / (2! × 2! × 2!) = 453600.
The word CALCULATOR consists of 4 vowels A, U, A, O. Let us consider them as a single letter say P. Therefore, now we have 7 letters: P, C, L, C, L, T, R in which “C” is repeated two times, and L is repeated two times. The number of arrangements of these 7 letters is 7! / (2! × 2!) = 1260.
After this is done, 4 vowels (in which ‘A’ is repeated 2 times) can be arranged in 4! / 2! = 12 ways.
Therefore, the number of arrangements of the letters of the word CALCULATOR in which vowels are together = 1260 × 12 = 15120.

Example 3:  John owns six-coloured pairs of shoes (two red, two blue and two black). He wants to put all these pairs of shoes on the shoe rack. How many different arrangements of shoes are possible?

Solution:

The total number of pairs of shoes = n = 6
Number of red shoes = p = 2
Number of blue shoes = q = 2
Number of black shoes = r = 2
This is an example of permutation with repetition because the elements are repeated and their order is important.
Put the above values in the formula below to get the number of permutations:
P(n; p, q, r) = n! / (p! × q! × r!)
P(6; 2, 2, 2) = 6! / (2! × 2! × 2!) = 720 / 8 = 90
Hence, shoes can be arranged on the shoe rack in 90 ways.

Example 4: A person has to choose three digits from the set of the following seven numbers to make a three-digit number. {1, 2, 3, 4, 5, 6, 7} How many different arrangements of the digits are possible?

Solution:

A three-digit number can have 2 or three identical numbers. Similarly, in a number, the order of digits is important. It is given that the person can select 3 digits from the set of 7 numbers. Hence, n = 7 and k = 3.
Substitute these values in the formula below to get the number of possible arrangements.
P(n, k) = n^k
P(7, 3) = 7^3 = 343
Hence, 343 different arrangements are possible.

Example 5:

If there are 4 chocolate chips, 2 oatmeal, and 2 double chocolate cookies in a box, in how many different orders is it possible to eat all of these cookies?

Solution:

Total number of cookies = 8, so n = 8.
We want to eat all of the cookies; therefore, we are choosing 8 objects at a time. Therefore, r = 8.
In the group of 8 cookies, there are 4 chocolate chips, 2 oatmeal, and 2 double chocolate cookies.
We are eating a group of cookies with flavours that repeat.
x1 = 4, x2 = 2, x3 = 2
nPr / (x1! × x2! × x3!) = 8P8 / (4! × 2! × 2!) = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1 × 2 × 1 × 2 × 1) = 40320 / 96 = 420

Example 6: In how many ways can the alphabets of the word EXCELLENT be arranged?

Solution: 

Total number of elements in the word = n = 9
E is repeated three times, hence p = 3
L is repeated 2 times, hence q = 2
Substitute these values in the formula below to get the number of ways in which the letters of this word can be arranged:
P(n; p, q) = n! / (p! × q!)
P(9; 3, 2) = 9! / (3! × 2!) = 362880 / 12 = 30240
Hence, the letters in the word EXCELLENT can be arranged in 30240 ways.

Example 7: Consider arranging the letters of the word "MATH". We want to find the number of distinct permutations possible.

Solution:

Since all the letters in the word "MATH" are distinct, we can directly apply the permutation formula without considering repetition.
Using the formula for permutations without repetition:
P = n! / (n1! × n2! × … × nk!)
In this case, since all the letters are distinct, we have:
P = 4! / (1! × 1! × 1! × 1!)
P = 24 / (1 × 1 × 1 × 1) = 24
Therefore, there are 24 distinct permutations of the letters in the word "MATH".

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