Posted by M.A. on Tuesday, February 9, 2010 at 9:45am.
The correct answer is nine.
Student A can get any of three books: b, c or d. Assume he gets b. Then there are only three possibilities of distributing the remaining three (a, c and d) among students B, C and D. B can get a, c or d, but there is only one way to distribute the other two. The same logic applies to each of the three possible assignments to student A.
3 x 3 = 9
hey how come u didn't mind giving the correct answer this time drwls????
The hardest part of the problem for me was figuring out a method of solution. My first attempt, using factorial numbers, was wrong. So I took more time to think about it, and even made a list of possible rearragements. Computing the number was easy, once the logic was done properly.
Students who are only looking for a number to plug into a computerized test sheet are missing the fun of mathematics, and will probably never master the subject.
Let the students be abcd
and the exams ABCD
Here are the possible ways to distribute exams:
BCDA
BDAC
BADC
CADC
CDAB
CDBA
DABC
DCAB
DCBA
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