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# 数学代写|组合学代写Combinatorics代考|Recursions

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## 数学代写|组合学代写Combinatorics代考|Recursions

Let’s explore yet another approach to evaluating the binomial coefficient $C(n, k)$. As in the previous section, let $S=\left{x_1, \ldots, x_n\right}$. We’ll think of $C(n, k)$ as counting $k$-subsets of $S$. Either the element $x_n$ is in our subset or it is not. The cases where it is in the subset are all formed by taking the various $(k-1)$-subsets of $S-\left{x_n\right}$ and adding $x_n$ to them. The cases where it is not in the subset are all formed by taking the various $k$-subsets of $S-\left{x_n\right}$. What we’ve done is describe how to build $k$-subsets of $S$ from certain subsets of $S-\left{x_n\right}$. Since this gives each subset exactly once,
$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{l} n-1 \ k-1 \end{array}\right)+\left(\begin{array}{c} n-1 \ k \end{array}\right)$$
by the Rule of Sum.
The equation $C(n, k)=C(n-1, k-1)+C(n-1, k)$ is called a recursion because it tells how to compute $C(n, k)$ from values of the function with smaller arguments. This is a common approach which we can state in general form as follows.

Technique. Deriving recursions Answering the question “How can I construct the things I want to count by using the same type of things of a smaller size?” usually gives a recursion.
Sometimes it is easier to answer the question “How can I break the things I want to count up into smaller things of the same type?” This usually gives a recursion when it is turned around to answer the previous question.

## 数学代写|组合学代写Combinatorics代考|Multisets

Let $M(n, k)$ be the number of ways to choose $k$ elements from an $n$-set when repetition is allowed and order doesn’t matter. Will any of our three methods for handling $C(n, k)$ work for $M(n, k)$ ? Let’s examine them.

• Imposing an order: The critical observation for our first method was that an unordered list can be ordered in $k$ ! ways. This is not true if repetitions are allowed. To see this, note that the extreme case of $k$ repetitions of one element has only one ordering.
• Using a recursion: We might be able to obtain a recursion, but we would still be faced with the problem of solving it.
• Using generating functions: To use the generating functions we have to allow for repetitions. This can be done very easily: Simply replace $\left(1+x_i\right)$ in Example 1.14 (p. 19) with the infinite sum
$$1+x_i+x_i^2+x_i^3+\cdots,$$
a geometric series which sums to $\left(1-x_i\right)^{-1}$. Why does this replacement work? When we studied $C(n, k)$ in Example 1.14, the two terms in the factor $1+x_i$ corresponded to not choosing the $i$ th element OR choosing it, respectively. Now we need more terms: $x_i x_i$ for when the $i$ th element is chosen to appear twice in our unordered list, $x_i x_i x_i$ for three appearances, and so forth. The distributive law still takes care of producing all possible combinations. As in Example 1.14, if we replace $x_i$ by $x$ for all $i$, the coefficient of $x^k$ will be the number of multisets of size $k$. Thus $M(n, k)$ is the coefficient of $x^k$ in $(1-x)^{-n}$. You should be able to use this fact and Taylor’s Theorem to obtain $M(n, k)=(n+k-1) ! /(n-1) ! k !$.

## 数学代写|组合学代写Combinatorics代考|Labelled versus unlabelled enumeration

$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{l} n-1 \ k-1 \end{array}\right)+\left(\begin{array}{c} n-1 \ k \end{array}\right)$$

## 数学代写|组合学代写Combinatorics代考|Surjections and set partitions

$$1+x_i+x_i^2+x_i^3+\cdots,$$

## MATLAB代写

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