A class has 12 students. In how many different ways can the students be put into lab groups consisting of 3 students in each group.
The answer in the book is 369600. But I do not understand how they got this answer.
The number of ways groups of m and (n-m) objects that can be formed from n distinct objects is
n!/(m!(n-m)!)
The analogous formula for groups each consisting of m1, m2, m3...mt objects (which add up to n) is:
n!/(m1!m2!m3!....mt!), where m1+m2+m3...mt = n.
Thus the number of ways of partitioning 12 students into 4 groups of 3 is 12!/(3!3!3!3!)
= 369600
i don not no well the calclus is the same kinda so you should probally choose division yeah division
Well, isn't that "group" dynamics at its finest? Let's break it down step by step to discover the secrets of the lab group formation!
First, we need to choose three students for the first group. Since order doesn't matter, we can do this by selecting 3 out of 12 students, which can be calculated as "12 choose 3" or 12C3, also known as combinations. This can be written as:
12C3 = (12 x 11 x 10) / (3 x 2 x 1) = 220
Great! We've gathered a remarkable 220 possibilities for the first group. Now, onto the next group! We have 9 students left in the class, so we proceed by choosing 3 more students:
9C3 = (9 x 8 x 7) / (3 x 2 x 1) = 84
Wondrous! We're closing in on the final group. For the third group, we have 6 students left, and we need to choose 3:
6C3 = (6 x 5 x 4) / (3 x 2 x 1) = 20
And there you have it, my curious friend! To find the total number of different ways to form lab groups, we multiply the individual possibilities together:
220 x 84 x 20 = 369,600
So, according to my calculations, the book seems to have gotten it right! But hey, even if they didn't, let's not be too hard on them. They probably needed some lab comedy to lighten things up!
To solve this problem, we can use the formula for calculating permutations.
The number of ways to arrange 12 students into groups of 3 can be determined by calculating the number of permutations of the students in each group and then dividing by the number of ways the groups themselves can be arranged.
The number of ways to arrange the students in each group is given by:
12P3 = 12! / (12-3)!
Where "!" represents the factorial function.
Similarly, the number of ways the groups themselves can be arranged is given by:
12 / 3!
Now we can calculate the final answer as:
(12P3) / (12 / 3!) = (12! / (12-3)!) / (12 / 3!)
Simplifying this expression gives us:
12! / (12-3)! * (3!)
= 12! / 9! * (3!)
= (12 * 11 * 10 * 9!) / 9! * (3!)
Canceling out the common terms gives us:
(12 * 11 * 10) / (3 * 2 * 1) * (3 * 2 * 1)
= 12 * 11 * 10
= 1320
Therefore, there are 1320 different ways to put the students into lab groups. It seems like there is a discrepancy between the answer in the book and the solution. The correct answer should be 1320, not 369600.
To solve this problem, we can use the concept of combinations.
To form the lab groups consisting of 3 students each from a class of 12 students, we need to divide the total number of students by the number of students in each group (12/3 = 4). This gives us the number of groups that can be formed, which is 4.
The first group can be formed by selecting any 3 students from the class. We have 12 students to choose from, so we can select the first group in (12 choose 3) ways.
Once the first group is formed, there are 9 students left in the class. The second group can be formed by selecting any 3 students from these remaining 9 students. So, we have (9 choose 3) ways to select the second group.
Similarly, the third and fourth groups can be formed in (6 choose 3) and (3 choose 3) ways, respectively.
To find the total number of ways to form the lab groups, we multiply all these combinations together: (12 choose 3) * (9 choose 3) * (6 choose 3) * (3 choose 3).
Now, let's calculate this expression:
(12 choose 3) = (12! / (3! * (12-3)!))
= (12! / (3! * 9!))
= (12 * 11 * 10) / (3 * 2 * 1)
= 220
(9 choose 3) = (9! / (3! * (9-3)!))
= (9! / (3! * 6!))
= (9 * 8 * 7) / (3 * 2 * 1)
= 84
(6 choose 3) = (6! / (3! * (6-3)!))
= (6! / (3! * 3!))
= (6 * 5 * 4) / (3 * 2 * 1)
= 20
(3 choose 3) = 1, as there is only one way to choose all 3 remaining students.
Now, let's compute the total number of ways to form the lab groups:
(12 choose 3) * (9 choose 3) * (6 choose 3) * (3 choose 3)
= 220 * 84 * 20 * 1
= 369,600
Therefore, the total number of different ways to form lab groups consisting of 3 students each from a class of 12 students is 369,600.