"Combinatorial Math" 04-Basic counting issues

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1. Basic Count 1.1 Unified model

This article discusses a few basic counting problems, although all have their own models, but intrinsically connected, so we first set up a unified model. Now that there are element sets \ (e,f\), their elements have intrinsic structure, build mappings \ (e\to f\), the question is how many of these mappings are there? How many are full-shot and single-shot?

The so-called internal structure, which is the topological structure between elements, the number of mappings we call, strictly speaking, is the number of equivalence classes in the topological isomorphic sense. There are a wide range of topologies that cannot be studied, this article only explores two basic topologies, and the next one will be more general discussion. Although this model has a limited role in this chapter, it can consider the nature of the problem from a higher perspective, and the links between the various issues are also at a glance.

The two basic topologies are: unordered structure and a linked list. The elements in an unordered structure are discrete and undifferentiated, and can be replaced with each other under topological isomorphism. The elements in the list are completely differentiated, its topological isomorphism only itself, of course, only its own topological structure of more than this one, as long as the emphasis on the complete distinction, so-called linked list can be used as the number of each element. Under \ (e,f\) are finite sets, and Kee \ (| e|=m,| f|=n\).

1.2 Model 1: distinguishable \ (\to\) distinguishable

Can completely distinguish the structure is relatively simple, first see \ (e,f\) are the case of the linked list, \ (e,f\) vertical arrangement, \ (E\to f\) is the general function definition. Each element of \ (e\) has a \ (n\) value that can be mapped, which is known by the multiplication principle as a common \ (n^m\) species. This structure has a more commonly used, more intuitive model that examines a set of \ (n\) letters (s\), using these letters to make up the words \ (x_1x_2\cdots x_m\) of length \ (m\). This word is also called \ (s\) on the \ (m\) meta- permutation , or \ (m\) meta-word . It is not difficult to prove that it and model 1 equivalence, so \ (s\) on the \ (m\) meta-word has \ (n^m\).

Now add some restrictions to the mapping, such as the Assumption \ (e\) \ (k\) can only take a \ (n_k\) value, by the multiplication theorem can have a \ (n_1n_2\cdots n_m\) \ (m\) meta-word. Again, the mapping of each element cannot be the same, or there are no duplicate letters in the word, and the k\ element can only fetch \ (n-k+1\) values. This is what we are familiar with \ (n\) elements of the (m\) elements of the permutation number (P (n,m) \) (formula (1)), where the expression \ (x (x-1) \cdots (x-k+1) \) précis-writers to \ ((x) _k\), also known as \ (x\) the descending \ (k\) factorial (\ (x\) does not require a natural number). Similarly, the expression \ (x (x+1) \cdots (x+k-1) \) Précis-writers is \ ((x) ^k\), also called \ (x\) the L \ (k\) factorial .

\[p (n,m) = (n) _m=n (n-1) \cdots (n-m+1) =\dfrac{n!} {(N-M)!} \tag{1}\]

The number of permutations requires that each function value is mapped at most once and continues to be generalized, we assume that \ (f\) Number of k\ is mapped \ (m_k\) times. Such a word is called the \ (a_1^{m_1}a_2^{m_2}\cdots a_n^{m_n}\) type Word, where \ (m_1+m_2+\cdots+m_n=m\). First the \ (m_k\) letters \ (a_k\) as the different letters \ (a_{k1},\cdots,a_{km_k}\), such words have \ (m!\), and the original each word corresponds to the \ (M_1!m_2!\cdots m_n!\) Such a word, Therefore, the formula (2) is established.

\[n (A_1^{m_1}a_2^{m_2}\cdots a_n^{m_n}) =\dfrac{m!} {M_1!m_2!\cdots m_n!} \tag{2}\]

   the number of \ (m\) characters on the \ (s\) that are not the same as the adjacent letters.

1.3 Model 2: indistinguishable \ (\to\) distinguishable

Now assume that \ (e\) is unordered, and \ (f\) is a linked list, because the original image is not distinguishable, it is to focus on the composition of the image. It's an equivalent model we're also familiar with: n\ (m\) a ball from a ball of color. If you require each color can only choose one, the equivalent of mapping to a single shot, select the number of \ (n\) of the \ (m\) element of the number of non-repetition , as \ (\binom{n}{m}\), generally referred to as the combination number. In fact, it is the number of equivalence classes arranged in the order of \ (e\), in particular, the m!\ arrangement belongs to the same equivalence class, so there is a formula (3).

\[\binom{n}{m}=\dfrac{(n) _m}{m!} =\dfrac{n!} {m! (N-M)!} \tag{3}\]

If the color can be repeated, select the number of \ (n\) of the \ (m\) meta-repeatable combination , as \ (\left (\binom{n}{m}\right) \). Set the number of times each image is mapped to \ (x_k\) and the number of non-negative integer solutions that are equivalent to the equation (4). There are two ingenious methods for this problem, they all transform the repeatable combinatorial problem into the non-repeating combinatorial problem, and the corresponding thought is very important. One method is to directly understand the equation (4), which is equivalent to dividing \ (m\) The same element into a distinguishable \ (n\) copy. The \ (m\) element is lined up, in which a \ (n-1\) partition can be, partitions and elements of the common \ (m+n-1\) seats, we are to choose the \ (n-1\) location, which is the formula (5).


\[\left (\binom{n}{m}\right) =\binom{m+n-1}{m}=\dfrac{(n) ^m}{m!} \tag{5}\]

Another interpretation (4) method is also very classical, the study of \ (S_k=x_1+\cdots+x_k\) composed of the monotone increment sequence (formula (6) left), \ (s_k\) in \ ([n]\) value, but there may be duplicate values. In order to eliminate the equal sign, then (t_k=s_k+k-1\), then a formula (6) is established, where \ (t_k\) in \ ([m+n-1]\) to take the reciprocal value, the same equation (5). Using this idea, it is more easy to get the inequality (7) Non-negative integer solution to \ (\binom{m+n}{m}\), it is also equivalent to the left-hand (0\sim m\) when the coefficient and, so the formula (8) is established.

\[0\leqslant s_1\leqslant\cdots\leqslant s_n=m\;\leftrightarrow\;0\leqslant t_1<\cdots<t_n=m+n-1,\;t_k=s_k+ K-1\tag{6}\]

\[x_1+x_2+\cdots+x_n\leqslant M\tag{7}\]


   \ ([n]\) take \ (m\) A number of non-adjacent, to find the number of the extraction.

1.4 Model 3: distinguishable \ (\to\) indistinguishable

As indistinguishable, it is like \ (n\) The same basket, the distinguishable image as a m\ different ball, the problem is equivalent to: \ (m\) a different ball into the \ (n\) Basket method. According to the number of baskets with balls \ (k\), the problem is divided into \ (n\) cases, each of which is the same problem: \ (m\) The division of a ball into a (k\) heap. We focus on this problem, and \ (m\) elements are divided into \ (k\) Part of the number is called the second class Stirling number , recorded as a \ (S (m,k) \).

There is no simple expression for the number of Stirling, we start with its recursive relationship, which is also a common method of group number calculation problem. Examine an element \ (a_1\), after splitting, there are two cases: it is separated into a bunch, or with other elements in a heap. So it's easy to have a recursive relationship (9). Using recursive relationship induction, you can get a more beautiful formula (10), if you think positive proof is not obvious, you can try reverse proof.

\[s (m,k) =s (m-1,k-1) +ks (m-1,k) \tag{9}\]

\[s (k+r,k) =\sum_{1\leqslant k_1\leqslant k_2\leqslant\cdots\leqslant k_r\leqslant k}k_1k_2\cdots k_r\tag{10}\]

Considering the relationship between Model 3 and model 1, the 1 splits in Model 3 are an equivalence class in Model 1: The number of images to be mapped is \ (k\). This equivalence class has \ (\binom{n}{k}k!= (n) _k\) elements (first selected \ (K\) as a permutation), so we have an important relationship (one) (\ (S (m,0) =0\)). \ (m\) The total number of split elements is called the Bell number , recorded as \ (b_m\), it is obviously the formula (12) left. It is easy to have a recursive relationship (12) to the right by classifying the size of the heap in which an element resides.

\[n^m=\sum_{k=0}^ns (M,k) (n) _k\tag{11}\]

\[b_m=\sum_{k=0}^ms (m,k) \;\rightarrow\; B_{m+1}=\sum_{k=0}^m\binom{m}{k}b_{m-k}\tag{12}\]

1.5 Model 4: indistinguishable \ (\to\) not distinguishable

When both the image and the image are indistinguishable, only the number of heaps and the number of each heap is concerned, and the equivalent model is the number of integers \ (m\) (p (n) \). The corresponding partition is the number of \ (k\) part of the (P (m,k) \), which represents the indefinite equation (13) The number of positive integer solutions, note that it differs from the equation (4) is not to consider the order of splitting. Since the order of each part is not considered, each partition actually corresponds to a set of nonnegative integer solutions of equation (14).

\[x_1+x_2+\cdots+x_k=m,\;\;(x_1\geqslant x_2\geqslant\cdots\geqslant x_k) \tag{13}\]


The split number also has no simple expression, so let's start by setting up some recursive relationships. The recursive relationship (15) is obtained based on whether \ (x_k\) is \ (1\), and the formula (16) is used continuously, and it has obvious combinatorial meanings.

\[p (m,k) =p (m-1,k-1) +p (m-k,k) \tag{15}\]

\[p (k+r,k) =p (r,1) +p (r,2) +\cdots+p (r,k) \tag{16}\]

With respect to a partition result of \ (m\) \ (x_1,x_2,\cdots,x_k\), there is an intuitive representation of an element that is not distinguishable, in the line \ (i\) lines (x_i\) points. For example, ((5,5,3,2) \) Painting, such a figure called Fierer (Ferrers), using its intuition can be a lot of unexpected conclusions. After the reversal of the Filetoux, the corresponding spin-off is called the conjugate spin -off of the original spin-off, and for the spin-off \ (\pi\), the conjugate spin-off is generally recorded as \ (\pi^*\). It is easy to get from the map: (1) \ (P (m,k) \) equals \ (m\) The maximum number of divisions for \ (k\), (2) \ (m\) The number of partitions (k\), equal to \ (m\) the largest division is not greater than \ (k\) of the number of partitions.

The \pi=\pi^*\ is called self-conjugate spin -offs, and its Filetoux is symmetrical diagonally. By dividing its Filetoux, it is known that the number of self-conjugate splits of m\ is equal to the number of different odd-numbered partitions of each part. Then see the different parts of the split, each section \ (x_k\) can be uniquely represented as \ (I\cdot 2^j\), where \ (i\) is odd. Split the section into \ (2^j\) \ (i\), and finally get all parts of the odd-numbered spin-off. The inverse of the various parts of the split, because the size of \ (I\) The number of parts can only be represented as \ (2\) The power of the second and, therefore, the corresponding relationship is bidirectional. That is to say, the number of separate parts is equal to the number of odd split.

2. Nature of the basic count 2.1 two-term coefficient

As we have discussed earlier, the number of combinations \ (\binom{n}{m}\) is the coefficients of the 1+x (x^m\) in the expansion of the \ ((() ^n\), and of course the coefficients of the \ (x^my^{n-m}\) items in the two-item (17), so it is also called a two-item coefficient . The definition of the number of combinations (3), \ (\dfrac{n!} {m! (N-M)!} \) requires \ (m,n\) to be a non-negative integer, but \ (\dfrac{(n) _m}{m!} \) in \ (n\) can be any real number, later form \ (\binom{x}{m}\) can also be used to denote a polynomial \ (\dfrac{(x) _m}{m!} \)。 This promotion is actually reasonable, and in calculus we know that \ ((1+x) ^{\alpha}\) of the Power series expansion type (18) shown (Newton two-term theorem), it is ((1+x) ^n\) promotion. In addition, the number of repeatable combinations can also be written as a formula (19).

\[(x+y) ^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}\tag{17}\]

\[(1+x) ^{\alpha}=1+\sum_{k=1}^{\infty}\binom{\alpha}{k}x^k\tag{18}\]

\[\left (\binom{n}{m}\right) = ( -1) ^m\binom{-n}{m}\tag{19}\]

In high school we know that some of the properties of the Hi Shang coefficients are not difficult to prove, here are just lists. In the formula (20), the symmetry and recursion are respectively, in which the recursive nature has obvious combinatorial meanings. Formula (21) is a single-peak of the change of m\. ((1+x) ^n\) (x=\pm 1\) available (22) and (23), in addition to the ((1+x) ^m (1+x) ^n= (1+x) ^{m+n}\) can get the van der Mong Heng equation (Vandermonde) (24).


\[\binom{n}{0}<\binom{n}{1}<\cdots<\binom{n}{\lfloor\frac{n}{2}\rfloor}=\binom{n}{\lceil\frac{n}{2}\ Rceil}>\cdots>\binom{n}{n}\tag{21}\]


\[\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\cdots=2^{n-1}\tag{23}\ ]


The two-term coefficients have been studied very early, and the most famous of course is the Yang Hui triangle (Pascal Triangle), which is using the formula (20) right. In the inversion formula that chapter we already know that The matrix \ (\{a_{ij}=\binom{i}{j}\}\) inverse matrix is \ (\{b_{ij}= ( -1) ^{i-j}\binom{i}{j}\}\), so there is a formula (25) Set up, of course, you can also directly prove it.

\[\sum_{k=0}^n ( -1) ^{k+m}\binom{n}{k}\binom{k}{m}=\delta_{mn}=\left\{\begin{matrix}1,&\text{if}\;m=n\\0, &\text{if}\;m\ne N\end{matrix}\right.\tag{25}\]

   proof: \ (\sum\limits_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}\);

   proof: \ (\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}=n\cdot 2^{n-1}\);

   write out the expansion of \ ((x_1+x_2+\cdots+x_p) ^n\).

2.2 Stirling number

Now we're going to talk about the second class of Stirling, and from the name we know it's a story number. The formula (10) is not satisfying, it is even just another description of the counting problem, only the formula (11) can be used. The formula obviously satisfies the form of a single chain \ (l_m\) inverse equation, which is arranged as formula (25) left (in order to get the correlation function \ (\binom{n}{k}\)), and the formula (25) right is obtained by the inverse equation, thus having another explicit expression (26) of \ (S (n,k) \).

\[n^m=\sum_{k=0}^n\binom{n}{k}k! S (m,k) \;\leftrightarrow\;n! S (m,n) =\sum_{k=0}^n ( -1) ^{n-k}\binom{n}{k}k^m\tag{25}\]

\[s (m,k) =\dfrac{1}{k!} \sum_{i=0}^k ( -1) ^{k-i}\binom{k}{i}k^m\tag{26}\]

Since the formula (11) is set to arbitrary \ (N\leqslant m\), the equation (27) is set to the left, thus \ (S (m,k) \) gives the linear table-out relationship of the polynomial group \ (\{x^m\}\) and \ (\{(x) _m\}\). Such a linear list must have an inverse relationship (formula (27) right), where \ (s (m,k) \) is called the first class Stirling number, which is related to the first class Stirling number (28).

\[x^m=\sum_{k=0}^ms (M,k) (x) _k\;\leftrightarrow\;(x) _m=\sum_{k=0}^ms (m,k) x^k\tag{27}\]

\[\sum_{k=0}^ms (m,k) s (k,n) =\delta_{nm}=\left\{\begin{matrix}1,&\text{if}\;m=n\\0,&\text{if}\;m\ne n\end {matrix}\right.\tag{28}\]

On the other hand, the investigation formula (27) Right, it is not difficult to know \ (s (m,k) \) There is a recursive (29), similarly there is an expression (30). \ (s (m,k) \) Positive and negative cross appear, in order to find its combined meaning, investigate the expression (31) in the \ (c (m,k) \). It is obviously the absolute value of \ (s (m,k) \), and the satisfying formula (32). Easy to verify, containing \ (k\) a rotation of the \ (m\) Yuan permutation number, also satisfies the formula (32) recursion relation, this is its combination meaning.

\[s (m,k) =s (m-1,k-1)-(m-1) s (m-1,k) \tag{29}\]

\[s (k+r,k) = ( -1) ^r\sum_{1\leqslant k_1\leqslant k_2\leqslant\cdots\leqslant k_r\leqslant k+r}k_1k_2\cdots k_r\tag{30 }\]

\[(x) ^M=\SUM\LIMITS_{K=0}^MC (m,k) x^k\tag{31}\]

\[c (m,k) =c (m-1,k-1) + (m-1) C (m-1,k) \tag{32}\]

• The number of 1^{\lambda_1}2^{\lambda_2}\cdots (m^{\lambda_m}\) type permutations.

"Combinatorial Math" 04-Basic counting issues

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