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# 统计代写|Story proofs stat代写

## 统计代考

A story proof is a proof by interpretation. For counting problems, this often means counting the same thing in two different ways, rather than doing tedious algebra. A story proof often avoids messy calculations and goes further than an algebraic proof toward explaining why the result is true. The word “story” has several meanings, some more mathematical than others, but a story proof (in the sense in which we’re using the term) is a fully valid mathematical proof. Here are some examples of story proofs, which also serve as further examples of counting.

Example 1.5.1 (Choosing the complement). For any nonnegative integers $n$ and $k$ with $k \leq n$, we have
$$\left(\begin{array}{l} n \ k \end{array}\right)=\left(\begin{array}{c} n \ n-k \end{array}\right)$$
This is easy to check algebraically (by writing the binomial coefficients in terms of factorials), but a story proof makes the result easier to understand intuitively.

Story proof: Consider choosing a committee of size $k$ in a group of $n$ people. We know that there are $\left(\begin{array}{l}n \ k\end{array}\right)$ possibilities. But another way to choose the committee is to specify which $n-k$ people are not on the committee; specifying who is on the committee determines who is not on the committee, and vice versa. So the two sides are equal, as they are two ways of counting the same thing.

Example 1.5.2 (The team captain). For any positive integers $n$ and $k$ with $k \leq n$,
$$n\left(\begin{array}{l} n-1 \ k-1 \end{array}\right)=k\left(\begin{array}{l} n \ k \end{array}\right)$$
This is again easy to check algebraically (using the fact that $m !=m(m-1) !$ for any positive integer $m$ ), but a story proof is more insightful.

Story proof: Consider a group of $n$ people, from which a team of $k$ will be chosen, one of whom will be the team captain. To specify a possibility, we could first choose the team captain and then choose the remaining $k-1$ team members; this gives the left-hand side. Equivalently, we could first choose the $k$ team members and then choose one of them to be captain; this gives the right-hand side.

Example 1.5.3 (Vandermonde’s identity). A famous relationship between binomial coefficients, called Vandermonde’s identity, ${ }^{2}$ says that
$$\left(\begin{array}{c} m+n \ k \end{array}\right)=\sum_{j=0}^{k}\left(\begin{array}{c} m \ j \end{array}\right)\left(\begin{array}{c} n \ k-j \end{array}\right)$$
This identity will come up several times in this book. Trying to prove it with a brute force expansion of all the binomial coefficients would be a nightmare. But a story proves the result elegantly and makes it clear why the identity holds.

Story proof: Consider a student organization consisting of $m$ juniors and $n$ seniors, from which a committee of size $k$ will be chosen. There are $\left(\begin{array}{c}m+n \ k\end{array}\right)$ possibilities. If there are $j$ juniors in the committee, then there must be $k-j$ seniors in the committee. The right-hand side of the identity sums up the cases for $j$.
Example 1.5.4 (Partnerships). Let’s use a story proof to show that
$$\frac{(2 n) !}{2^{n} \cdot n !}=(2 n-1)(2 n-3) \cdots 3 \cdot 1$$
Story proof: We will show that both sides count the number of ways to break $2 n$ people into $n$ partnerships. Take $2 n$ people, and give them ID numbers from 1 to $2 n$. We can form partnerships by lining up the people in some order and then saying the first two are a pair, the next two are a pair, etc. This overcounts by a factor of $n ! \cdot 2^{n}$ since the order of pairs doesn’t matter, nor does the order within each pair. Alternatively, count the number of possibilities by noting that there are $2 n-1$ choices for the partner of person 1 , then $2 n$ – 3 choices for person 2 (or person 3 , if person 2 was already paired to person 1), and so on.

## 统计代考

$$\left(\begin{数组}{l} n \ ķ \end{array}\right)=\left(\begin{array}{c} n \ n-k \end{数组}\右）$$

$$n\left(\begin{数组}{l} n-1 \ k-1 \end{array}\right)=k\left(\begin{array}{l} n \ ķ \end{数组}\右）$$

$$\left(\begin{数组}{c} m+n \ ķ \end{array}\right)=\sum_{j=0}^{k}\left(\begin{array}{c} 米\ j \end{array}\right)\left(\begin{array}{c} n \ k-j \end{数组}\右）$$

$$\frac{(2 n) !}{2^{n} \cdot n !}=(2 n-1)(2 n-3) \cdots 3 \cdot 1$$