Let's analyze and solve the questions:

1. Combien y a-t-il de 4-uplets d’éléments distincts de $E$?

An ensemble $E$ has 9 elements, and we need to form 4-uplets of distinct elements.

This involves permutations without repetition, as the order of the elements in the 4-uplet matters. The formula for permutations is:

$P(n, k) = \frac{n!}{(n-k)!}$

Here:

• $n = 9$ (the total number of elements),
• $k = 4$ (the number of elements to select).

Substituting:

$P(9, 4) = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024$

So, there are 3024 distinct 4-uplets.

2a. Combien de codes peut-il alors créer ?

Sven needs to create a 6-digit code with the following restrictions:

• He cannot use the same digit twice.
• The digit 0 cannot be used.

This leaves him with 9 available digits ($1, 2, \dots, 9$).

The number of possible permutations is again calculated using:

$P(n, k) = \frac{n!}{(n-k)!}$

Here:

• $n = 9$,
• $k = 6$.

Substituting:

$P(9, 6) = \frac{9!}{(9-6)!} = \frac{9!}{3!} = 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 60480$

Thus, Sven can create 60,480 codes.

2b. Combien de codes différents sont possibles si Jérémy a vu son ami taper 5 comme dernier chiffre ?

If the last digit of the code is fixed as $5$, Sven now only has to arrange the remaining $5$ digits from the set $\{1, 2, 3, 4, 6, 7, 8, 9\}$ (excluding $5$).

The number of permutations for $5$ digits out of $8$ is:

$P(8, 5) = \frac{8!}{(8-5)!} = \frac{8!}{3!} = 8 \times 7 \times 6 \times 5 \times 4 = 6720$

So, there are 6720 possible codes where $5$ is the last digit.

Final Answers:

1. 3024 4-uplets of elements distinct from $E$.
2. a. 60,480 codes Sven can create. b. 6720 codes if $5$ is the last digit.

Would you like a detailed explanation for any of the steps or further related examples? Here are 5 related questions:

1. What is the formula for combinations ($C(n, k)$) and how does it differ from permutations?
2. How would the result change if repetition of digits were allowed in question 2?
3. Can you explain factorials and their importance in counting problems?
4. How can we use permutations to solve real-life problems like scheduling or seating arrangements?
5. What if $E$ had 10 elements instead of 9? How would the results for 4-uplets change?

Tip: When solving counting problems, always clarify whether order matters (permutations) or does not matter (combinations).