Permutations and combinations: formulas and derivations

pull into

Definition:

Sort: Sort a specified number of elements from a specified number of elements; (consider the order of the elements)
Combination: taking out just the specified number of elements from a given number of elements; (regardless of the order of the elements)

The Principle of Addition & Multiplication

Addition Principle:

Completing a project can have$n$ (b) Category approach.$a_i(i\in[1,n])$ delegate$i$ The number of class methods. The total number of methods to accomplish this is$S=a_1+a_2+a_3+···+a_n$ A different kind of approach.

Multiplication Principle:

Completion of a project requires$n$ Steps.$a_i(i\in[1,n])$ delegate$i$ The number of different methods for each step. Then to complete the project there are a total of$S=a_1 \times a_2 \times a_3 \times ··· \times a_n$ A different kind of approach.

Fundamentals of permutations and combinations

Arrangement number:

through (a gap)$n$ Any one of the different elements of the$m(m \leqslant n, m and n are natural numbers, same below) $ The elements are arranged in a column in a certain order, called from$n$ Of the different elements$m$ An arrangement of the elements; from$n$ Remove from the different elements$m(m \le n)$ The number of all permutations of an element is called the number of permutations from the$n$ Remove elements from the$m$ The number of permutations of the elements, symbolized by$A^m_n$ (or$P^m_n$ ) indicates.
The formula for the alignment is as follows:

$A^m_n =n(n-1)(n-2)..(n-m+1)=\frac{n!}{(n-m)!}$

$n!$ in the name of$n$ The factorial of$6!=1 \times 2 \times 3 \times 4 \times 5 \times 6$ 。

The formula can be interpreted like this:$n$ candidate$m$ The captain of the team is$n$ ( $m \leqslant n$ ), the first position can be chosen by the person who$n$ The second position can be selected by$n-1$ and so on, the first$m$ (last one) can be selected$n-m+1$ One, got to:

\[A^n_m =n(n-1)(n-2)...(n-m+1)= \frac{n!}{(n-m)!} \]

Full alignment:$n$ Individuals to line up, with a captain for$n$ The first position can be selected. The first position is optional.$n$ One, the second position can be selected$n-1$ and so on:

\[A^n_n=n(n-1)(n-2)...3 \times 2 \times 1=n! \]

Full permutations are a special case of permutation numbers.

Number of combinations:

through (a gap)$n$ Of the different elements, take any$m(m \leqslant n)$ elements to form a set called from$n$ Remove from the different elements$m$ A combination of elements; from$n$ q out of the different elements$m(m \leqslant n)$ The number of all combinations of elements is called the number of combinations from the$n$ Remove from the different elements$m$ The number of combinations of elements, symbolized by$(^n_m)$ To represent this, read $\left\lceil n elect m \right\rfloor $ .
Combination number formula:

\[( ^n_m )=\frac{A^m_n}{m!}=\frac{n!}{m!(n-m)!} \]

How to understand the above formula? Consider$n$ candidate$m$ (after a verb) indicating that a person is not aware of it$(m \leqslant n)$ , no queue, don't care about order. If one cares about the order, it is$A^m_n$ If you don't care then you have to get rid of the duplicates, so how many duplicates are there? The same.$m$ Individuals, but also $ \left\lceil full alignment \right\rfloor $ get$m!$ So get:

\[( ^n_m )\times m!=A^m_n \]

\[(^n_m )=\frac{A^m_n}{m!}=\frac{n!}{m!(n-m)!} \]

Combinations are also commonly used$\complement^m_n$ denote$\complement^m_n=(^n_m)$ 。

The combinatorial numbers are also known as $ \left\lceil binomial coefficients \right\rfloor $ , the connection of which is described below.

Special, when$m>n$ when$A^m_n=(^n_m)=0$ 。

system of inserting plates

The plugboard method is a technique used to find the number of solutions to a class of schemes that group the same elements, and can also be used to find the number of groups of solutions to a class of linear equations.

The topic of sums of positive integers

$Q_1$ :: Existing$n$ The requirement to categorize exactly the same elements into$k$ How many ways are there to divide the groups, ensuring that each group has at least one element?

For this problem we can abstractly understand that in the$n-1$ Insertion in the gaps$k-1$ The entire queue is divided into$n$ part, two partitions cannot be adjacent. This makes it a classical combinatorial number problem. It is possible to obtain

\[ans=\complement^{k-1}_{n-1}=\frac{{(n-1)}!}{{(k-1)}!(n-k)!} \]

Its essence is to seek$x_1+x_2+x_3+...+x_k=n$ The number of sets of positive integer solutions to the

The topic of sums of non-negative integers

$Q_2$ :: What if it is allowed to be empty?

At this point it is not possible to insert the boards, because there may be many boards inserted into a null case, which is very bad for computation. Therefore, we consider constraining it so that it is transformed into a restricted$Q_1$ . first of all, I'd like to borrow$k$ elements, which are evenly distributed to the$k$ Group, in$n+k$ elemental$n+k-1$ For an air-lipped panel, the

\[ans=\complement^{k-1}_{n+k-1}=\complement^{n}_{n+k-1} \]

Since the elements are identical, taking one from each group after it has been divided into groups has no effect at all on the result, i.e., the results are equal.
Its essence is to seek$x_1+x_2+x_3+...+x_k=n(x_i\geqslant 0)$ The number of groups of non-negative integer solutions of the

The topic of sums of integers with different lower bounds

$Q_3$ : If each group has at least$t$ What about an element?

At this point, for the inset method, this leads to an increase in the spacing of the plates, which is not well calculated. Same$Q_2$ , and see how that translates into$Q_1$ . First, the$t-1$ elements are pressed into the groups, it translates into the number of elements in the$n-k(t-1)-1 (in particular, define n-1>k(t-1))$ Insert boards in the gaps.

\[ans=\complement^{k-1}_{n-k(t-1)-1} \]

disjoint alignment

exist$[1,n]$ get a position by passing the imperial exam$k$ This.$k$ The combination of any two numbers that are not adjacent to each other is given by the fact that there are$\complement^{k}_{n-k-1}$ Kind.

the Binomial Theorem (math.)

The Binomial Theorem clarifies the coefficients of an expansion:

\[(a+b)^n= \sum\limits_{i=0}^n (^n_i)a^{n-i}b^i \]

Use of mathematical induction:

Acting it out by hand will reveal that for${(a+b)}^n$ (coll.) fail (a student)$n=0$ when it is equal to$1$ , and so on, respectively

For a moment:${(a+b)^1}=a+b$
For the two times:${(a+b)}^2=(a+b)(a+b)=a^2+2ab+b^2$
For three times:${(a+b)^3}=(a+b){(a+b)}^2=a^3+3a^2b+3ab^3+b^3$

Now you can look for patterns.

We will write it like a tower:

Now just look at the coefficients:

Isn't that very familiar?
Yes, this is the Yang Hui Triangle:

A graphically explicit presentation of the Binomial Theorem, Yanghui's triangle presents us with a property of combinations, viz:

\[\complement^{k-1}_{n}+\complement^{k}_{n}=\complement^{k}_{n+1} \]

Not only that, but for$(a+b)^n$ The coefficients of the expansion of the Yang-Hui triangle correspond in turn to the first$n+1$ each term in the row (Binomial Theorem).

a binomial coefficient (math.)

Binomial coefficients can be arranged to form Pascal's triangle (Yang Hui triangle). If the binomial coefficients are arranged in a row, in a top-down order, they form a Pascal's triangle.

Binomial coefficients are commonly found in various fields of mathematics, especially combinatorial mathematics. In fact, its can be understood as the coefficients from$n$ Selection of different elements$k$ The number of methods for the number of elements. The definition of binomial coefficients can be generalized to$n$ is the case of complex numbers and is still referred to as a binomial coefficient.