Math (combination/permutation)
posted by Cynthia .
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. 
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.