Question 1:
There are 12 students in a class. Find the number of ways that 12 studenets can take 3 different tests if 4 students are to take each test.
Question 2
Find the number of ways 12 students can be partitioned into 3 teams A, B, C, so that each team contain 4 students.
I don't get what is the difference between the two. For the first one, the book uses C(12,4) * C(8,4) = 34650. I use the permutation method, which is bascially the same 12!/(4!4!4!) = 34650
For the second question, the books tells me to think that A is in team A and you have to chose three different people to join team A, resulting into C(12,3). The same logic applies to team B: C(7,3) and that left 4 for team C.
Therefore the answer was C(12,3) * C(7,3) = 5925.
THe book tells me if you use the partition method, all you do is 34650/6 = 5925 because 3! is an ordered partition. (what is an ordered partition anyway?)
What is the difference between partition and combination?
I see no difference in the two problems.
You are simply breaking up 12 students into 3 groups of 4, the fact that they are taking a test is irrelevant.
the first group can be formed in C(12,4) ways, leaving
8 students to form the second group which is C(8,4).
That leaves 4 to form the last group of C(4,4)
total number of way is
C(12,4) x C(8,4) x C(4,4)
If you break it down by definition this is
12!/(4!8) x 8!/(4!4!) x 4!/(4!0!) or 12!/(4!4!4!) which is your answer.
Your permutation answers looks like the formula of arranging 12 things, of which 4 are alike of one kind, 4 of another, and 4 of yet another.
Your reasoning is correct, you are seeing each student in a group as "alike"
If the book tells you that "you have to chose three different people to join team A, resulting into C(12,3)" they are wrong, since each team is to contain 4 students, not 3.
Did you check your typing?
I don't see any connection of the calculation of 34650/6 with the problem.
Well, if you want to understand the difference between partition and combination, let me explain it in a humorous way:
Imagine you have a delicious pizza with 12 slices, and you want to share it among your 3 hungry friends. You have two options:
Option 1 (Combination): Each friend gets to choose their own slices independently, without any specific order. So, you decide to let your friends each take 4 slices. You can calculate this using combinations. The result would be C(12,4) * C(8,4) = 34650. It's like saying, "Here, take 4 slices, and you take 4, and you take 4!"
Option 2 (Partition): Now, you want to assign each friend to a specific team, like Team A, Team B, and Team C. You don't care about the order in which they are assigned to the teams. Using partitions, you take the total number of combinations from Option 1 (34650) and divide it by the number of ways you can arrange the teams, which is 3! (3 factorial). The result would be 34650/6 = 5925. It's like saying, "Hey, you three, join Team A. You three, join Team B. And you three, join Team C!"
In a nutshell, combination focuses on the number of ways you can choose elements without considering order, while partition takes into account the order in which elements are assigned to different groups or teams.
So, whether you're dividing pizza slices or grouping students, understanding the difference between partition and combination can help you share the fun while keeping things organized!
Let A denote one of the students. There are C(11,3) = 165 ways to choose 3 other students to be on the same tam as A. Now let B a student who is not on the same as A. Then let B be a studnt who is not on the same team as A. There are C(7,3) = 35 ways to choose 3 from the remaining students to be on the same team as b. The remaining 4 students form the third team. Thus the answer is 35*165 = 5925
Above is the exact wording.
The book solution undercounts because there are 12 ways to fix the first member of A, and 8 ways to fix the first member of B.
However, computing 12*C(11,3)*8*C(7,3) overcounts by making the first member of A and B distinctive.
The correct solution is the same as the first problem.
The first question is asking for the number of ways to distribute 12 students into 3 different tests, with 4 students taking each test. This is a combination problem because the order in which the students are selected for a particular test does not matter.
The book's solution uses the combination formula: C(n, r) = n! / (r!(n-r)!), where n is the total number of students and r is the number of students selected for each test. In this case, C(12,4) represents the number of ways to select 4 students out of 12 and C(8,4) represents the number of ways to select 4 students out of the remaining 8 after the first test. Multiplying these two combinations together gives the total number of ways to distribute the students to the tests.
Alternatively, you used the permutation formula: P(n, r) = n! / (n-r)!, which counts the number of ways to arrange a certain number of objects in a specific order. However, this formula is not applicable to this question because the order of selection does not matter.
For the second question, the difference lies in the interpretation of the distribution. Instead of distributing students into different tests, you are partitioning them into different teams. This is also a combination problem because the order of team selection does not matter.
The book's solution uses the combination method again. For team A, you need to select 4 students out of the 12, which is represented by C(12,3). Similarly, for team B, you need to select 4 students out of the remaining 7 (as 3 have already been selected for team A), which is represented by C(7,3). Multiplying these two combinations together gives the total number of ways to partition the students into teams.
The book mentions an ordered partition, which means that the teams A, B, C are distinct and have a specific order. In this context, an ordered partition is a permutation of the teams. Therefore, dividing the total number of ways by 3! (which is 6, as there are 3 teams) adjusts the count to account for the ordered nature of the partition.
In summary, the difference between partition and combination lies in the interpretation of the distribution. In the first question, the students are being distributed into different tests, while in the second question, they are being partitioned into different teams. Both problems can be solved using combination formulas, but the second question requires adjustment for an ordered partition if necessary.