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# 数学代写|离散数学代写Discrete Mathematics代考|MATH271 The Binomial Theorem

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## 数学代写|离散数学代写Discrete Mathematics代考|The Binomial Theorem

In Chapter 1 we introduced the numbers $\left(\begin{array}{l}n \ k\end{array}\right)$ and called them binomial coefficients. It is time to explain this strange name: it comes from a very important formula in algebra involving them, which we discuss next.

The issue is to compute powers of the simple algebraic expression $(x+y)$. We start with small examples:
\begin{aligned} (x+y)^2 &=x^2+2 x y+y^2 \ (x+y)^3 &=(x+y) \cdot(x+y)^2=(x+y) \cdot\left(x^2+2 x y+y^2\right) \ &=x^3+3 x^2 y+3 x y^2+y^3 \end{aligned}
and continuing like this,
$$(x+y)^4=(x+y) \cdot(x+y)^3=x^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4 .$$

## 数学代写|离散数学代写Discrete Mathematics代考|Distributing Presents

Suppose we have $n$ different presents, which we want to distribute to $k$ children, where for some reason, we are told how many presents each child should get. So Adam should get $n_{\text {Adam }}$ presents, Barbara, $n_{\text {Barbara }}$ presents, etc. In a mathematically convenient (though not very friendly) way, we call the children $1,2, \ldots, k$; thus we are given the numbers (nonnegative integers) $n_1, n_2, \ldots, n_k$. We assume that $n_1+n_2+\cdots+n_k=n$, else there is no way to distribute all the presents and give each child the right number of them.

The question is, of course, how many ways can these presents be distributed?

We can organize the distribution of presents as follows. We lay out the presents in a single row of length $n$. The first child comes and takes the first $n_1$ presents, starting from the left. Then the second comes and takes the next $n_2$; then the third takes the next $n_3$ presents etc. Child $k$ gets the last $n_k$ presents.

It is clear that we can determine who gets what by choosing the order in which the presents are laid out. There are $n$ ! ways to order the presents. But of course, the number $n$ ! overcounts the number of ways to distribute the presents, since many of these orderings lead to the same results (that is, every child gets the same set of presents). The question is, how many?
So let us start with a given distribution of presents, and let’s ask the children to lay out the presents for us, nicely in a row, starting with the first child, then continuing with the second, third, etc. This way we get back one possible ordering that leads to the current distribution. The first child can lay out his presents in $n_1$ ! possible orders; no matter which order he chooses, the second child can lay out her presents in $n_2$ ! possible ways, etc. So the number of ways the presents can be laid out (given the distribution of the presents to the children) is a product of factorials:
$$n_{1} ! \cdot n_{2} ! \cdots n_{k} !$$
Thus the number of ways of distributing the presents is
$$\frac{n !}{n_{1} ! n_{2} ! \cdots n_{k} !}$$

## 数学代写|离散数学代写Discrete Mathematics代考|The Binomial Theorem

$$(x+y)^2=x^2+2 x y+y^2(x+y)^3 \quad=(x+y) \cdot(x+y)^2=(x+y) \cdot\left(x^2+2 x y+y^2\right)=x^3+3 x^2 y+3 x y^2+y^3$$

$$(x+y)^4=(x+y) \cdot(x+y)^3=x^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4 .$$

## 数学代写|离散数学代写Discrete Mathematics代考|Distributing Presents

$$n_{1} ! \cdot n_{2} ! \cdots n_{k} !$$

$$\frac{n !}{n_{1} ! n_{2} ! \cdots n_{k} !}$$

## MATLAB代写

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