Question 6864: I am stuck on a question from my text book, on Inequalities, Permutations, and Probibility. Can someone help me? My question is....A teacher has a set of 12 problems to use on a math exam. The teacher makes different versions of the exam by putting 10 questions on each exam. How many different exams can the teacher make. I am not sure how to set up this problem. I keep multiplying 10 and 12 but I don't think that is right. Can someone help me. My paper is due tomorrow.
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! i often get stuck on permutations and combinations myself, so i always end up going to first principles by thinking of an easier solution.
Got to remember that both perms and combos use factorials either straightforward factorials or factorials divided by factorials, so how about having 4 questions and picking 3? How many? answer is ABC, ABD, ACD, BCD -->4.
So, 4 questions is 4! --> 120. Now we need to make this smaller, by division: dividing by 3! would give an answer of 4.
4 questions: pick 1 is 4 possible combinations, 4!/3!. Actually, by thinking about 1 question from 4 or 2 from 4, the actual calc is 4!/(3!1!).
so, your question is 12!/(10!2!) = 66... basically:
1234567890
123456789A
123456789B
123456780A
123456780B
etc etc
Hope this makes sense.
jon.
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