Question

Труппа состоит из 10 артистов. Сколькими способами можно выбрать из нее в течение двух вечеров по 6 человек для участия в спектаклях так, чтобы эти составы не совпадали друг с другом

Answer 1

Если я не ошибаюсь то только 1 сплсбом

Answer 2

Problem Analysis

We are given a theater troupe consisting of 10 artists, and we need to determine the number of ways we can select two different groups of 6 artists each over the course of two evenings for participation in performances, such that the two groups do not overlap.

Solution

To solve this problem, we can use the concept of combinations. The number of ways to choose a group of 6 artists from a total of 10 is given by the combination formula:

C(n, r) = n! / (r! * (n-r)!)

Where n is the total number of artists and r is the number of artists we want to choose.

Since we need to select two different groups of 6 artists each, we can calculate the number of ways for each evening separately and then multiply them together.

Calculation

Let's calculate the number of ways for each evening:

For the first evening: - n = 10 (total number of artists) - r = 6 (number of artists to choose)

Using the combination formula, we can calculate the number of ways to choose a group of 6 artists from a total of 10 for the first evening:

C(10, 6) = 10! / (6! * (10-6)!)

For the second evening, we need to choose another group of 6 artists, but we need to ensure that this group does not overlap with the first group. This means we have 4 remaining artists to choose from.

For the second evening: - n = 4 (remaining number of artists) - r = 6 (number of artists to choose)

Using the combination formula, we can calculate the number of ways to choose a group of 6 artists from a total of 4 for the second evening:

C(4, 6) = 4! / (6! * (4-6)!)

Finally, we can multiply the number of ways for each evening together to get the total number of ways to choose two different groups of 6 artists each:

Total number of ways = C(10, 6) * C(4, 6)

Let's calculate this:

C(10, 6) = 10! / (6! * (10-6)!) = 210

C(4, 6) = 4! / (6! * (4-6)!) = 0 (since we cannot choose 6 artists from a pool of 4)

Total number of ways = 210 * 0 = 0

Therefore, there are 0 ways to choose two different groups of 6 artists each over the course of two evenings, such that the two groups do not overlap.

Answer

There are 0 ways to choose two different groups of 6 artists each over the course of two evenings, such that the two groups do not overlap.