The natural numbers from 1 to 50 (inclusive) are written on the board. How many integers must be deleted at a minimum to ensure that for all remaining integers $x$ and $y$, $x+y$ is not a prime number?

## Solutions
Let $m$ denote the minimum we are looking for.

We start by noting that by removing the 25 odd (or even) integers, all the sums of the remaining integers are even and $\geqslant 2$: they are therefore not prime numbers, and the property is verified.

Thus, $m \leqslant 25$.

Now, consider pairs of integers that cannot remain together:

- $\{46,1\},\{45,2\}, \ldots,\{24,23\}$ because all their sums are 47, which is prime.
- $\{50,47\},\{49,48\}$ because $50+47=48+49=97$, which is prime.

By the pigeonhole principle, if it were possible to delete strictly fewer than 25 integers, by the pigeonhole principle, two integers from the same pair would not have been deleted, but their sum would be a prime number, which is contradictory.

It follows that $m \geqslant 25$ and therefore $m=25$.

## 3 Second Part: Algebra and Geometry

## 1 Counting (Jean)

The goal of this session is to focus on counting calculations, particularly binomial coefficients.

## Products, Sums, and Factorials

We start by briefly recalling the first principles of counting.

The first essential principle of combinatorics is that for making successive and independent choices, we multiply the number of possibilities.

## Example 1.

In a deck of cards, if there are 13 possibilities for the value and 4 possibilities for the suit, then there are $13 \times 4=52$ possible cards.

Remark 2. From a set theory perspective, this means that the cardinality of a Cartesian product is the product of the cardinalities.

If $E$ and $F$ are finite sets, then

$$
|E \times F|=|\{(e, f), e \in E, f \in F\}|=|E| \times|F|
$$

The second principle is that to count a disjoint union in pairwise disjoint cases, we sum the possibilities in each case.

Remark 3. From a set theory perspective, this means that the cardinality of a disjoint union is the sum of the cardinalities.

If $A$ and $B$ are finite sets and $A \cap B=\emptyset