To solve the problems in the uploaded image, I'll first transcribe and translate the exercises:

Exercise 1.5:

How many ways can 8 students be seated at a round table?

This is a problem involving circular permutations. In circular permutations, the number of ways to arrange $n$ objects in a circle is given by: $(n - 1)!$ For 8 students, we have: $(8 - 1)! = 7! = 5040$ So, there are 5040 ways to seat 8 students at a round table.

Exercise 1.6:

How many different rearrangements of the word "COSMOS" are possible?

The word "COSMOS" contains 6 letters in total, but some letters are repeated:

• C occurs 1 time,
• O occurs 2 times,
• S occurs 2 times,
• M occurs 1 time.

The formula for counting the number of distinct permutations when some objects are identical is: $\frac{n!}{k_1!k_2!\dots k_r!}$ where $n$ is the total number of objects, and $k_1, k_2, \dots, k_r$ are the frequencies of the repeated objects.

For "COSMOS", the number of rearrangements is: $\frac{6!}{2!2!} = \frac{720}{4} = 180$ So, there are 180 different rearrangements of the word "COSMOS".

Exercise 1.7:

How many different rearrangements of the word "LOKOS" are possible?

The word "LOKOS" contains 5 letters, with:

• L occurs 1 time,
• O occurs 2 times,
• K occurs 1 time,
• S occurs 1 time.

The number of distinct permutations is: $\frac{5!}{2!} = \frac{120}{2} = 60$ So, there are 60 different rearrangements of the word "LOKOS".

Exercise 1.8:

How many ways can the letters of the word "COMBINATORICS" be rearranged?

The word "COMBINATORICS" contains 13 letters, with:

• C occurs 2 times,
• O occurs 1 time,
• M occurs 1 time,
• B occurs 1 time,
• I occurs 2 times,
• N occurs 1 time,
• A occurs 1 time,
• T occurs 1 time,
• R occurs 1 time,
• S occurs 1 time.

The number of distinct permutations is: $\frac{13!}{2!2!} = \frac{6227020800}{4} = 1556755200$ So, there are 1,556,755,200 ways to rearrange the letters of the word "COMBINATORICS".

Exercise 1.11:

How many ways can 28 guests be seated at two round tables, with 14 guests at each table?

This is a combination of two circular permutations. The number of ways to arrange 14 guests at a round table is: $(14 - 1)! = 13!$ For two tables, there are two groups of 14, so the number of arrangements is: $(13! \times 13!) \div 2!$ We divide by $2!$ to account for the fact that the two tables are indistinguishable.

Thus, the total number of ways to seat 28 guests at two round tables is: $\frac{(13!)^2}{2} = \frac{(6,227,020,800)^2}{2} = 19,433,008,009,600 \div 2 = 9,716,504,004,800$ So, there are 9,716,504,004,800 ways to seat 28 guests at two round tables.

Would you like further details or have any questions on the explanations?

Here are five related questions to extend your understanding:

1. How would the seating arrangements change if two students switch places at the round table?
2. What is the difference in permutations if there were more repeated letters in "COSMOS"?
3. How would you calculate the number of rearrangements if the word "LOKOS" had more than two repeated letters?
4. What are some applications of circular permutations in real-world problems?
5. How would you calculate seating arrangements for groups of different sizes at multiple round tables?

Tip: When calculating permutations for circular objects, always remember to subtract 1 from the total number to account for the rotational symmetry of the arrangement.