# 数学代写|组合学代写Combinatorics代考|CS519

## 数学代写|组合学代写Combinatorics代考|Properties of Binomial Coefficients

1. For any numbers $n$ and $k(n \geq 1 ; k=0,1, \ldots, n)$, the following equality holds
$$C_n^{n-k}=C_n^k .$$
This equality means that in the sequence
$$C_n^0, C_n^1, C_n^2, \ldots, C_n^{n-1}, C_n^n$$
the numbers equidistant from its ends are the same:
$$C_n^n=C_n^0, \quad C_n^{n-1}=C_n^1, \quad C_n^{n-2}=C_n^2, \ldots$$
Why?
First of all, this comes from the computational formula:
$$C_n^k=\frac{n !}{k !(n-k) !}, C_n^{n-k}=\frac{n !}{(n-k) !(n-(n-k)) !}=\frac{n !}{(n-k) ! k !} .$$
However, combinatorial proof featuring modeling is much more attractive and instructive.
Equality (3.25) evidence that an $n$-element set contains the same amount of $(n-k)$ element and $k$-element subsets. If we manage to verify this directly, without counting, then we will obtain another proof of equality. It is easy to guess that a bijection can be established between the complements of two types of subsets: we match those two subsets, namely $k$ element subset and $(n-k)$-element one, the union of which is the whole $n$-element set. Below, there is an example of such correspondence in the case $n=5, k=2$.

Let $M={a, b, c, d, e}$ be a five-element set. Below we present the pairs of two- and three-element sets matched by complements: 2-element and 3-element subsets

These two eolumns explieitly and unequivocally validate the equality
$$C_5^2=C_5^3 .$$
Construct similar columns and apply them to prove the equality $C_5^1=C_5^4$.

1. For any numbers $n$ and $k(n \geq 2 ; k=1,2, \ldots, n-1)$ the following equality holds:
$$C_n^k=C_{n-1}^{k-1}+C_{n-1}^k .$$

## 数学代写|组合学代写Combinatorics代考|Raising Binomials to Powers. Newton’s Binomial Formula

It may seem that we have always known the formula of raising the sum of two terms to the power of 2 :
$$(a+b)^2=a^2+2 a b+b^2 .$$
If necessary, we can recall the above formula easily and without any effort. It would seem quite surprised if one used the chain of transformations
$$(a+b)^2=(a+b) \cdot(a+b)=a \cdot a+a \cdot b+b \cdot a+b \cdot b=a^2+2 a b+b^2 .$$
every time, when it is necessary to get the right-hand side of equality from its left-hand side. We can reproduce the above steps but we are used to skipping the steps involved. The summands in the right-hand side, namely, $a^2, 2 a b$ and $b^2$, have reliably imprinted in our memory.

The above concerns to the formula of raising the sum of two terms to the power 3 as well:
$$(a+b)^3=a^3+3 a^2 b+3 a b^2+b^3 .$$
When required, we can easily deduce a similar formula for the fourth power of the above sum. Here is this formula:
\begin{aligned} &(a+b)^4=(a+b)^3 \cdot(a+b)= \ &=\left(a^3+3 a^2 b+3 a b^2+b^3\right) \cdot(a+b)= \ &=a^4+4 a^3 b+6 a^2 b^2+4 a b^3+b^4 . \end{aligned}
Basing on (3.38), it is straightforward to derive the formula for the fifth power of binomial and so on. We strongly suggest the reader take a while to develop similar formulas for the fifth and sixth powers.

It does not make much sense to spawn this type of formula for the higher powers (seventh, eighth, etc.). It is much more feasible (and interesting) to find out if the rules (3.36), (3.37), (3.38), etc., which are applicable in the special cases of raising of a binomial to the second, third, fourth, etc. power can be generalized for the case of power $n$.

Guessing the General Rule. In order to detect the common patterns of the formulas of raising of a binomial to powers $2,3,4,5,6$, we need to write down the list of these formulas in compact form. It is also useful to accompany them with the obvious formula for the case of power 1 . The resulting list of formulas is presented below.

# 组合学代写

## 数学代写|组合学代写Combinatorics代考|Properties of Binomial Coefficients

1. 对于任何数字 $n$ 和 $k(n \geq 1 ; k=0,1, \ldots, n)$, 以下等式成立
$$C_n^{n-k}=C_n^k .$$
这个相等意味着在序列中
$$C_n^0, C_n^1, C_n^2, \ldots, C_n^{n-1}, C_n^n$$
与其末端等距的数字是相同的:
$$C_n^n=C_n^0, \quad C_n^{n-1}=C_n^1, \quad C_n^{n-2}=C_n^2, \ldots$$
为什么?
首先，这来自计算公式:
$$C_n^k=\frac{n !}{k !(n-k) !}, C_n^{n-k}=\frac{n !}{(n-k) !(n-(n-k)) !}=\frac{n !}{(n-k) ! k !} .$$
然而，以建模为特征的组合证明更具吸引力和指导性。
平等 (3.25) 证明一个 $n$-元溸集包含相同数量的 $(n-k)$ 元嫊和 $k$-元素子集。如果我们设法直接验证 这一点，不计算，那么我们将获得另一个相等的证明。很容易猜到可以在两类子集的补集之间建立双 射：我们匹配这两个子集，即 $k$ 元嫊子集和 $(n-k)$-元青一，它的并集是整体 $n$-元素集。下面，有一 个案例中这种对应的例子 $n=5, k=2$.
让 $M=a, b, c, d, e$ 是一个五元綁集。下面我们介绍由补码匹配的二元和三元集合对：二元子集和三元子 集
这两个 eolumns 明确且毫不含楜地验证了相等性
$$C_5^2=C_5^3 .$$
构造相似的列并应用它们来证明相等性 $C_5^1=C_5^4$.
2. 对于任何数字 $n$ 和 $k(n \geq 2 ; k=1,2, \ldots, n-1)$ 以下等式成立:
$$C_n^k=C_{n-1}^{k-1}+C_{n-1}^k .$$

## 数学代写|组合学代写Combinatorics代考|Raising Binomials to Powers. Newton’s Binomial Formula

$$(a+b)^2=a^2+2 a b+b^2 .$$

$$(a+b)^2=(a+b) \cdot(a+b)=a \cdot a+a \cdot b+b \cdot a+b \cdot b=a^2+2 a b+b^2 .$$

$$(a+b)^3=a^3+3 a^2 b+3 a b^2+b^3 .$$

$$(a+b)^4=(a+b)^3 \cdot(a+b)=\quad=\left(a^3+3 a^2 b+3 a b^2+b^3\right) \cdot(a+b)==a^4+4 a^3 b+6 a^2 b^2+4 a b^3$$

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