Counting Answer Keys
Date: 05/25/2001 at 12:52:38 From: Alison Vickery Subject: Combinations / Permutations A multiple-choice test has 30 questions, each with five choices. How many answer keys are possible? This is a question on my students' pre-algebra test. I believe the answer should be 5^30. The book claims the answer is 150 (or 5x30). Many of my students agree. I'm not sure. Can you help?
Date: 05/25/2001 at 16:22:25 From: Doctor Douglas Subject: Re: Combinations / Permutations Hi Alison, and thanks for writing. You are correct. If there is a single correct answer for each of the thirty questions, then there are 5^30 possible answer keys. It's instructive to check one's reasoning by seeing what happens if we change the problem to vary the number of questions, testing some small values for which we can actually list the possibilities: 1 question: clearly, 5 choices: A B C D or E By your reasoning, this is 5^1 = 5. By the book's reasoning, this is 5x1 = 5, so both formulas give the same result, and a test consisting of only one question does not distinguish between the two formulas. So we move on to a two-question test: 2 questions: AA, AB, AC, AD, AE, BA, BB, BC, BD, BE, CA, CB, CC, CD, CE, DA, DB, DC, DD, DE, EA, EB, EC, ED, EE for a total of twenty-five choices. By the (erroneous) reasoning in the book, we should have 2x5 = 10 answers, but we can enumerate twenty-five of them, so that the book's reasoning is clearly wrong. Your (correct) reasoning yields 5^2 = 25. For 3 questions, it would take me a long time to type out all 5^3 = 125 possibilities, but it's already clear that the formula 5x3 = 15 is going to be incorrect, since we can generate twenty-five different choices simply by preceding each of the twenty-five two-question keys (such as AB) by the letter A: AAA, AAB, AAC, AAD, AAE, ABA,..., AED, AEE. The other one hundred keys are formed when those twenty-five two-question keys are preceded by B, C, D and E. I hope this helps you and your students understand why we must use powers (rather than a single multiplication) for this type of counting problem. You might also consider asking your students the following question as you explain these concepts: On a thirty-question test, each of which has five answers, a student answers only ONE question. In how many different ways could this student have done so? Answer: 5x30 = 150. A list of these 150 possibilities looks like this: 1A, 1B, 1C, 1D, 1E, 2A, 2B, ... , 30C, 30D, 30E. Your students should readily agree that 150 is the correct answer to this problem, and now it becomes clear that 150 seriously undercounts the number of answer keys in the original problem above. I hope this helps. Please write back if you need more explanation about this! - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
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