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Stu Dent Errors Again!


Stu Dent solved the following problem incorrectly.

Problem: In how many ways can five dice be tossed so that at least three ones show? Explain.

Stu Dent's Solution: 5C362 = 360 ways.

Stu Dent's Reasoning: Three dice must show 1's and the other two dice may or may not show 1's.

First, what is the error in Stu's reasoning? Second, is it possible to obtain a correct answer by adjusting his reasoning process?

 

Source: Mathematics Magazine, 1956, pp. 50.


Hint: First, convince yourself that Stu's reasoning and answer are incorrect.

Then, other options are available...writing out the cases, looking for patterns, re-examining his use of 5C362.

 


Solution Commentary: Emil Schell (a Remington Rand Univac employee in 1956) submitted this "clear" response: "The student counted the tosses (1,1,1,x,y) as each occurring 5C3 = 10 times. Actually (1,1,1,1,1), for example, occurs only once. He may correct his solution by counting the permutations of (1,1,1,x,y), Consider these in three categories:

  • CASE I: x = y = 1. There is only one example. This occurs once but his solution counts it as occurring 10 times, giving a surplus of 9.
  • CASE II: Either x or y is one, and the other is not. There are five examples. Each occurs five times, but his solution counts each as occurring twenty times, giving a surplus of 75.
  • CASE III: Neither x nor y is one. There are thirty cases. These are correctly counted. Thus, 5C362 - 9 - 75 = 276."