Disposizioni semplici e permutazioni

• This summarization discusses simple arrangements and permutations including repeated arrangements.

Simple arrangements and permutations

• Considers the problem of arranging 10 contestants into a ranking for the top 3 positions.
• For the first place, there are 10 possibilities, for the second place, there are 9, and for the third place, there are 8.
• The total number of possible rankings is calculated using the fundamental principle of combinatorial calculus multiplying 10 * 9 * 8
• Classifying this arrangement as an ordered sequence of 3 athletes chosen from 10, without repetition
• Generalizes the problem to: Given n objects, what are the possible ordered sequences of k objects, chosen from the n assigned, without repeating objects?
• These types of sequences are specifically named

Simple arrangements

• With n distinct objects, a simple arrangement (or simply arrangement) of the n objects in k places, with k ≤ n, is any ordered sequence of k objects chosen from those assigned, without repeating objects.
• The number of arrangements of n objects in k places is also called the arrangements of n objects of class k.

Arrangements Theorem

• The total number of arrangements of n objects in k places, indicated by the symbol Dₙ,ₖ, is given by the formula: Dₙ,ₖ = n(n-1)(n-2)...(n-k+1)

Permutation

• Any ordering of n distinct objects is called a permutation
• From the arrangement formula, it follows that the number of permutations of n objects, indicated by the symbol Pₙ, is given by: Pₙ = Dₙ,ₙ = n(n-1)(n-2)...321
• The product n(n-1)(n-2)...321 is also indicated with the symbol n!, which is called "factorial of n" or "n factorial".
• The symbol n! remains defined for each positive integer n, while for n = 0, by definition, 0! = 1.

Permutation Theorem

• The total number of permutations of n distinct objects is given by the formula: Pₙ = n!
• For example: 3! = 3 * 2 * 1 = 6 and 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040

Arrangements with repetition

• Discusses the question of how ways it is possible to fill a column of the totocalcio, where each column consists of 14 boxes, each of which must be filled with one of the three symbols 1, 2, or X.
• For the first box, there are 3 possibilities of choice (1, 2, X), for the second box, we can still choose from 3 possibilities, and so on, until the fourteenth box, giving 3^14 total combinations
• The difference of this problem compared to those considered in the previous paragraph is that here the objects (the symbols 1, 2, X) can be chosen more than once, that is, they can be repeated.
• More generally the total possible sequences of k objects with n options with repetition allowed is considered

Arrangements Definition with Repetition

• Given n distinct objects, an arrangement with repetition of the n objects in k places is any ordered sequence of k objects, chosen from those assigned, allowing for the objects to be repeated.
• Since it is allowed to repeat the objects, k can be greater than or equal to n.

Arrangement Theorem with Repetition

• The total number of arrangements with repetition of n objects in k places, indicated with the symbol D'ₙ,ₖ, is given by the formula: D'ₙ,ₖ = n * n * ... * n = nᵏ, with k factors equal to n

Example Problems using Repetition

• The last seven digits, after the initial 347, can each be any number between 0 and 9; there are ten possibilities for each digit.
• There are as many cell phone numbers as there are arrangements with repetition of 10 objects in 7 places, i.e., 10⁷.
• If a test consists of 10 multiple-choice questions: A, B, C, or D, the number of ways it can be filled out is 4¹⁰ = 1048576.

Permutations with Repetitions

• Considers finding all anagrams of the word "mamma," noting it differs from "cielo" because letters repeat
• The general problem: How many possible permutations of n objects exist if they are not all distinct?

Definition

• Permutation with repetition is defined as any permutation of n objects where not all of them are distinct
• The number of possible permutations with repetition are calculated in order to obtain the total number, say x, of the anagrams of the word "mamma"
• The difference with all the same letter would be a factor of 3! for the different combinations (e.g. m1, m2, m3)
• Distinguishing now with an index also the 2 letters «а»: ma₁m2m3a2, the factor will be x 3! 2!

Repetition Theorem

• Given n objects, of which a₁ are equal to each other, a₂ are equal to each other, and with a₁+ a₂ + … + aₖ = n, the distinct permutations of these n objects are: n! / (a₁! a₂! ... ak!)

Examples using Repetition Theorem

• With "matematica", formed by 10 letters, but "m", "a" and "t" are repeated two to three times, while the 'e', 'c' and 'i' only one, the total is = 151200
• A way to fill the column is permuting with 14 elements which 4 match with 1, 6 match with X e 4 match with 2; the total ways to do it is = 2101210

Combinations

• This portion of the text constructs models focused on unordered arrangements
• A simple combination defines a possible arrangement
• Consider how many ways is it possible to make a combination of five numbers = terno? (three numbers correct)? = (1, 2, 3, 4, [5]), in how many can I find the correct terno.
• The terns aren't in order, must be considered a set
• They can't have repeated combination: {1, 2, 3}, (3, 4, 5), (2, 3, 5). Ai reggruppament di oggetti non ordinati senza ripetizioni si di un nome particolare.

Definition, Combination Theorem

• Given n elements, it is called an n level combination, that doesn't has order in the k objects with the connection to never repeat the object.
• A sequence is a combination representation ordered, where a permutation is a sequence that aren't ordered
• A sequence in this representation can't has combination, and they also can't overlap
• the elements not be repeated in consider order
• This part of the text identifies the number sub-level

Combining items with class n

• Is indicated with the symbol Cₙₖ , C
• The total number, in math, when the objects can't be repeated and we can't give order to a combination, we can identify with a sub-combination

What C

• What're the C n level? We're thinking with the first instance, with what it is proposed at the beginning, with the objective being the sub-combination {1, 2, 3, 4, 5);.
• It isn't possible to apply the next schema
• For C 5 3 can't be used to apply the first element but a sequence from a sub-combination
• The 3 chosen from the new create combination with difference order
• Each schema can be generalized
• The theorem to the next step is defined to the next lemma
• We can define the relation in sequence

The Number for Combinations

• C nk = formula 4 = the term will change
• It must match all combination in class k, and is the sub-combination
• = n level/(n-k)!k
• Each must be more accurate