$$
\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\ldots+(-1)^{k}\binom{n}{k}+\ldots+(-1)^{n}\binom{n}{n}=?
$$
As for the previous exercise, it involves recognizing a binomial of Newton but this time with $a=1$ and $b=-1$. We therefore find $(1-1)^{\mathfrak{n}}=0$.

## Corollary 1.14.

$$
\binom{n}{k}=\binom{n}{n-k}
$$

Proposition 1.15 (Pascal's Formula).

$$
\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}
$$

Proof. We want to choose $k$ elements from $n$. We isolate one element. Either this element is part of the chosen ones, or it is not. In the first case, we still need to choose $k-1$ elements from the remaining $n-1$, and in the second case, we need to choose $k$ elements from the remaining $n-1$. Thus,

$$
\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}
$$

Corollary 1.16. To find the value of $\binom{n}{k}$ for small $n$, we write Pascal's triangle:

![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-134.jpg?height=317&width=447&top_left_y=221&top_left_x=796)

It is sufficient to write "1" on the sides of the triangle, then each number is equal to the sum of the two numbers above it, according to the previous proposition. This easily provides the values of the binomial coefficients.

## 2 Algebra TD

## - Combinatorics -