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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/   
    
Associated Topics:
High School Permutations and Combinations

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