No results were found for the search term: At a student conference 12 students were asked to form different committees with the following conditions:1)each committee was to contain exactly 3 students;2)each student was to belong to exactly 2 committees;3)no two commmittees could have more tha one student in common(A and B could not be together on 2 different committees) How many committees were formed? m
12*2=24
24/3=8
8 committees
To solve this problem, we can use the principle of combinatorics.
Let's break down the given conditions:
1) Each committee is to contain exactly 3 students.
2) Each student is to belong to exactly 2 committees.
3) No two committees can have more than one student in common.
To find the number of committees formed, we need to determine the total number of ways we can choose 3 students from the available pool of students, while satisfying the given conditions.
Let's denote the total number of committees formed as 'm'.
To find 'm', we need to calculate the number of ways each student is assigned to the committees. Since each student belongs to exactly 2 committees, we can calculate this by using combinations.
The formula for the number of combinations is nCk = n! / (k!(n-k)!), where n is the total number of students and k is the number of students chosen for each committee.
In this case, n = 12 (total number of students) and k = 3 (number of students per committee).
Using the combination formula, we calculate the number of ways to choose 3 students out of 12 as:
C(12, 3) = 12! / (3!(12-3)!) = 12! / (3!9!) = (12 * 11 * 10) / (3 * 2 * 1) = 220
However, we need to consider the third condition that no two committees can have more than one student in common. This means that if a student is chosen for one committee, they cannot be chosen for another committee that has a common member.
To account for this, we divide the result obtained above by 2, because each student is only assigned to 2 committees:
m = 220 / 2 = 110
Therefore, a total of 110 committees were formed at the student conference.