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# 数学代写|组合学代写Combinatorics代考|The Inclusion-Exclusion Principle

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## 数学代写|组合学代写Combinatorics代考|The Inclusion-Exclusion Principle

Alternatively, we could make an indirect count by observing that the number of permutations of ${1,2, \ldots, n}$ with 1 in the first position is the same as the number $(n-1)$ ! of permutations of ${2,3, \ldots, n}$. Since the total number of permutations of ${1,2, \ldots, n}$ is $n !$, the number of permutations of ${1,2, \ldots, n}$ in which 1 is not in the first position is $n !-(n-1) !=(n-1) !(n-1)$.

Example. Count the number of integers between 1 and 600 , inclusive, which are not divisible by 6 .

We can do this indirectly as follows. The number of integers between 1 and 600 which are divisible by 6 is $600 / 6=100$ since every sixth integer is divisible by 6 . Hence $600-100=500$ of the integers between 1 and 600 are not divisible by 6 .

The rule used to obtain an indirect count in these examples is the following. If $A$ is a subset of a set $S$, then the number of objects in $A$ equals the number of objects in $S$ minus the number not in $A$. Recall that
$$\bar{A}=S-A={x: x \text { in } S \text { but } x \text { not in } A}$$
is the complement of $A$ in $S$-that is, the set consisting of those objects in $S$ which are not in $A$. The rule can then be written as
$$|A|=|S|-|\bar{A}| \text { or, equivalently, }|\bar{A}|=|S|-|A| \text {. }$$
This formula is the simplest instance of the inclusion-exclusion principle.

## 数学代写|组合学代写Combinatorics代考|Combinations with Repetition

In Sections 3.3 and 3.5 we have shown that the number of $r$-combinations of a set of $n$ distinct elements is
$$\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{r !(n-r) !}$$
and that the number of $r$-combinations of a multiset with $k$ distinct objects, each with an infinite repetition number, equals
$$\left(\begin{array}{c} r+k-1 \ r \end{array}\right) \text {. }$$
In this section we show how the latter formula, in conjunction with the inclusion-exclusion principle, gives a method for finding the number of $r$-combinations of a multiset without any restrictions on its repetition numbers.

Suppose $T$ is a multiset and an object $x$ of $T$ of a certain type has a repetition number that is greater than $r$. Then the number of $r$ combinations of $T$ equals the number of $r$-combinations of the multiset obtained from $T$ by replacing the repetition number of $x$ by $r$. This is so because the number of times $x$ can be used in an $r$-combination of $T$ cannot exceed $r$. Therefore, any repetition number that is greater than $r$ can be replaced by $r$. For example, the number of 8-combinations of the multiset ${3 \cdot a, \infty \cdot b, 6 \cdot c, 10 \cdot d, \infty \cdot e}$ is the same as the number of $8-$ combinations of the multiset ${3 \cdot a, 8 \cdot b, 6 \cdot c, 8 \cdot d, 8 \cdot e}$. We can summarize by saying that we have determined the number of $r$-combinations of a multiset $T=\left{n_1 \cdot a_1, n_2 \cdot a_2, \ldots, n_k \cdot a_k\right}$ in the two “extreme” cases:
(i) $n_1=n_2=\cdots=n_k=1$; (i.e., $T$ is a set) and
(ii) $n_1=n_2=\cdots=n_k=r$.

## 数学代写|组合学代写Combinatorics代考|The Inclusion-Exclusion Principle

$$\bar{A}=S-A={x: x \text { in } S \text { but } x \text { not in } A}$$

$$|A|=|S|-|\bar{A}| \text { or, equivalently, }|\bar{A}|=|S|-|A| \text {. }$$

## 数学代写|组合学代写Combinatorics代考|Combinations with Repetition

$$\left(\begin{array}{l} n \ r \end{array}\right)=\frac{n !}{r !(n-r) !}$$

$$\left(\begin{array}{c} r+k-1 \ r \end{array}\right) \text {. }$$

(i) $n_1=n_2=\cdots=n_k=1$;(例如，$T$是一个集合)和
(ii) $n_1=n_2=\cdots=n_k=r$。

## MATLAB代写

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