Question

Из 6 учеников 10 класса и 8 учеников 11 класса нужно составить комиссию из трех человек . Сколькими способами это можно сделать,если в комиссию должен войти не более,чем один десятиклассник?

Answer 1

Объясняю ( не для того, кто задал вопрос, а для тех, "кто в танке")
1)комиссия состоит из 3-х человек.
2) в комиссию может войти
     а) один из 6-ти десятиклассников и 2 из 8-и одиннадцатиклассников
     б) ни одного десятиклассника (т.к. понятие не более - это значит равно и меньше. Для людей - это 1 либо 0). Тогда в комиссии будут только 3 одиннадцатиклассника.
Решаем
       а) 2 из 8 одиннадцатиклассников = 8!/(2!*(8-2)!) =28   но на каждого из 6 десятикл. приходится 28 комбинаций из 2-х одиннадцатикл. , соответственно комиссию можно составить 28*6=168 способов.
       б) 3 из 8 одиннадцатикл. = 8!/(3!*(8-3)!)=56

т.е. всего возможных комбинаций при заданном условии задачи будет 168+56=224 способа.

Answer 2

Problem Analysis

We need to determine the number of ways to form a committee of three members from a group of six students in the 10th grade and eight students in the 11th grade, with the condition that there should be no more than one student from the 10th grade in the committee.

Solution

To solve this problem, we can consider two cases: when there is no student from the 10th grade in the committee and when there is exactly one student from the 10th grade in the committee.

# Case 1: No student from the 10th grade in the committee

In this case, we need to select all three members from the eight students in the 11th grade. The number of ways to do this is given by the combination formula:

Number of ways = C(8, 3)

Using the combination formula, we can calculate this as:

Number of ways = 8! / (3! * (8 - 3)!) = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56

# Case 2: Exactly one student from the 10th grade in the committee

In this case, we need to select one student from the six students in the 10th grade and two students from the eight students in the 11th grade. The number of ways to do this is given by the product of the number of ways to select one student from the 10th grade and the number of ways to select two students from the 11th grade:

Number of ways = C(6, 1) * C(8, 2)

Using the combination formula, we can calculate this as:

**Number of ways = (6! / (1! *

Answer 3

Problem Analysis

We need to determine the number of ways to form a committee of three members from a group of six students in the 10th grade and eight students in the 11th grade, with the condition that there should be no more than one student from the 10th grade in the committee.

Solution

To solve this problem, we can consider two cases: when there is no student from the 10th grade in the committee and when there is exactly one student from the 10th grade in the committee.

# Case 1: No student from the 10th grade in the committee

In this case, we need to select all three members from the eight students in the 11th grade. The number of ways to do this is given by the combination formula:

Number of ways = C(8, 3)

Using the combination formula, we can calculate this as:

Number of ways = 8! / (3! * (8-3)!) = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56

Therefore, there are 56 ways to form the committee when no student from the 10th grade is included.

# Case 2: Exactly one student from the 10th grade in the committee

In this case, we need to select one student from the six students in the 10th grade and two students from the eight students in the 11th grade. The number of ways to do this is given by the product of the number of ways to select one student from the 10th grade and the number of ways to select two students from the 11th grade:

Number of ways = C(6, 1) * C(8, 2)

Using the combination formula, we can calculate this as:

Number of ways = (6! / (1! * (6-1)!)) * (8! / (2! * (8-2)!)) = (6 * 8 * 7) / (1 * 2) = 168

Therefore, there are 168 ways to form the committee when exactly one student from the 10th grade is included.

# Total number of ways

To find the total number of ways to form the committee, we need to add the number of ways from both cases:

Total number of ways = Number of ways (Case 1) + Number of ways (Case 2) = 56 + 168 = 224

Therefore, there are 224 ways to form the committee from the given group of students, with the condition that there should be no more than one student from the 10th grade in the committee.

Answer

There are 224 ways to form the committee of three members from a group of six students in the 10th grade and eight students in the 11th grade, with the condition that there should be no more than one student from the 10th grade in the committee.