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: _{5}C_{3}6^{2} = 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, reexamining his use of _{5}C_{3}6^{2}.
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 _{5}C_{3} = 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, _{5}C_{3}6^{2}  9  75 = 276."
