Permutations and Combinations of a Multiset

When dealing with a multiset, the principles of permutations and combinations changes and formula for these also change. Unlike traditional sets, multisets allow for duplications, which significantly impacts the way we calculate arrangements and selections.

Imagine you are organizing a gift bag for a party, and you have a selection of candies to include. However, instead of each candy being unique, you have several of the same type. For instance, your multiset of candies includes:

• 3 identical chocolates
• 2 identical lollipops
• 1 gummy bear

This problem is solved using Permutations and Combinations of a Multiset which is discussed in this article.

Multiset Definition

A multiset is a mathematical structure which can contain one or many members of one element while a set contains one element only in case of the same set. For example, {a, a, b} we can note that it is a multiset because the element a repeats itself. The difference between multisets and sets is useful in counting problems where the repetition of elements should be considered in determining arrangement or selections.

Examples of Multisets

• {1, 1, 2} – A multiset where the element 1 is repeated.
• {a, a, a, b, b} – A multiset with three occurrences of a and two of b.
• {red, red, blue, blue, blue} – A multiset of colors, where "red" appears twice and "blue" appears three times.

Permutations of a Multiset

Permutations of a multiset refer to the number of distinct arrangements of elements in a multiset (a set where duplicates are allowed). A multiset permutation takes into account that identical items should not be treated as distinct, which is different from a regular set permutation.

Formula for Permutations of a Multiset

To calculate the number of distinct permutations of a multiset, you can use the formula:

P = \frac{n!}{k_1! \times k_2! \times \cdots \times k_r!}

Where:

• n is the total number of elements in the multiset.
• k₁, k₂, . . . ,k_r are the frequencies of each distinct element in the multiset.

Worked Example: Permutation of a Multiset

Example: Consider the multiset {A, A, B, B, C} and find all permutation of this multiset.

Solution:

The total number of elements n = 5.

The repetitions are:

• A appears 2 times.
• B appears 2 times.
• C appears 1 time.

The formula becomes:

P = \frac{5!}{2! \times 2! \times 1!} = \frac{120}{2 \times 2 \times 1} = \frac{120}{4} = 30

So, there are 30 distinct permutations of the multiset {A, A, B, B, C}.

Combinations of a Multiset

Combinations of a multiset refer to the selection of elements from a multiset where the order of selection does not matter, and repetitions are allowed. When dealing with multisets, we often encounter two types of combinations: combinations without repetition (like choosing elements from a set without considering how often they occur) and combinations with repetition (where you can choose an element multiple times).

Formula for Combinations of a Multiset

The formula for combinations with repetition is:

C = \binom{n + r - 1}{r} = \frac{(n + r - 1)!}{r! \cdot (n - 1)!}

Where n is the number of distinct elements in the multiset, and r is the number of elements to be chosen. This formula allows for the selection of repeated elements.

Worked Example: Combination of a Multiset

Example: Consider a multiset {A, B, C}, and you want to choose 3 elements (with repetition allowed).

Solution:

Using the formula:

\binom{3 + 3 - 1}{3} = \binom{5}{3} = \frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10

So, there are 10 ways to choose 3 elements with repetition allowed.

The possible combinations (including repetition) are:

{A, A, A}, {A, A, B}, {A, A, C}, {A, B, B}, {A, B, C}, {A, C, C}, {B, B, B}, {B, B, C}, {B, C, C}, {C, C, C}.

Special Cases and Advanced Applications

Multisets can be used in more advanced counting problems, where inclusion-exclusion principles or generating functions help manage constraints. These methods can tackle problems involving multisets with restrictions or conditions on how elements can be selected or arranged.

Inclusion-Exclusion Principle for Multisets

The counting principle such as inclusion-exclusion is also referred to as addition and subtraction of intersection of set sizes whereby one is capable of determining the number of elements in the union of some sets. This formula is very handy especially when you are working with multisets that meet more than one condition since the results have overlapping members.

Modified Formula for Multisets

For two sets, the inclusion-exclusion principle formula is:

|A_1 \cup A_2| = |A_1| + |A_2| - |A_1 \cap A_2|

For three sets, the formula extends to:

|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|

For n sets, the general inclusion-exclusion formula is:

|A_1 \cup A_2 \cup \dots \cup A_n| = \sum_{i} |A_i| - \sum_{i < j} |A_i \cap A_j| + \sum_{i < j < k} |A_i \cap A_j \cap A_k| - \dots + (-1)^{n+1} |A_1 \cap A_2 \cap \dots \cap A_n|

Where:

• A_i represents the sets,
• The first sum adds the sizes of all individual sets,
• The second sum subtracts the sizes of all pairwise intersections,
• The third sum adds the sizes of all triple intersections, and so on,
• This alternating addition and subtraction continue based on how many sets overlap.

Solved Questions on Permutations and Combinations of a Multiset

Question 1: Find the number of distinct permutations of the multiset {A, A, B, B, C}.

Solution:

The total number of elements is 5, but we have repetitions:

• A appears 2 times.
• B appears 2 times.
• C appears 1 time.

We use the formula for permutations of a multiset:

P = \frac{n!}{k_1! \times k_2! \times \cdots \times k_r!}

Where:

• n = 5 (total elements),
• k₁ = 2 (for A),
• k₂ = 2 (for B),
• k₃ = 1 (for C).

Thus, P = \frac{5!}{2! \times 2! \times 1!} = \frac{120}{2 \times 2 \times 1} = \frac{120}{4} = 30

There are 30 distinct permutations of the multiset {A, A, B, B, C}.

Question 2: From the multiset {A, A, B, C}, how many distinct combinations of 2 elements can be selected?

Solution:

We list all possible combinations of 2 elements:

{A, A}, {A, B}, {A, C}, {B, C}

Since combinations don’t consider order, these are all distinct combinations.

There are 4 distinct combinations: {A, A}, {A, B}, {A, C}, {B, C}.

Question 3: From the multiset {A, B, C}, how many distinct ways can you choose 3 elements, allowing repetitions?

Solution:

We use the formula for combinations with repetition:

\binom{n + r 1}{r} = \frac{(n + r 1)!}{r! \times (n 1)!}

Where:

• n = 3 (number of distinct elements: A, B, C),
• r = 3 (we are choosing 3 elements).

Substituting values:

\binom{3 + 3 1}{3} = \binom{5}{3} = \frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10

The distinct combinations are:

{A, A, A}, {A, A, B}, {A, A, C}, {A, B, B}, {A, B, C}, {A, C, C}, {B, B, B}, {B, B, C}, {B, C, C}, {C, C, C}.

There are 10 distinct combinations.

Practice Questions

Que 1: Find the number of distinct permutations of the multiset {P, P, Q, Q, Q, R}.

Que 2: How many distinct combinations of 3 elements can be selected from the multiset {X, X, Y, Z, Z}?

Que 3: Find the number of distinct permutations of the multiset {A, A, A, A, B, B, C, C}.

Que 4: From the multiset {A, B, C, D}, how many distinct ways can you choose 4 elements, allowing repetitions?

Que 5: How many distinct combinations of 2 elements can be selected from the multiset {M, M, M, N, N}?

Answer Key

1. 60
2. 5
3. 420
4. 35
5. 3

Conclusion

Permutations and combinations of multisets provide powerful tools for solving problems involving repeated elements. Whether arranging objects or selecting groups, the inclusion of repetition adds complexity that can be managed through clear formulas and principles. It is important for students working under the field of probability and statistics to have a good grasp on this concept.

Read More,

• Permutation and Combination
• Permutation
• Permutation Formula
• Restricted Permutation
• Circular Permutation
• Combination
• Combination Formula