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 teh 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.
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