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Mathematics 241 \hspace{9cm} \= {\bf Name:} Evan Harrington \\
March 24, 2017 \>  {\bf Homework 18}
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\noindent {\bf Problem 1:}
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    \item I decided to split this problem up by class. This student's schedule will consist of a total of 6 courses, meaning that there will be exactly $6!$ different ways that the schedule can be arranged assuming that the classes have been chosen. Since there are three different math courses and we must take two, there are 6 different sets of math classes to take. With 2 different Literature courses to take, there are 2 different sets of literature classes. Once we get into the math courses, it becomes harder. Four available math courses for two slots leads us to 12 different sets of two classes. Lastly, there are 7 different sets of physical education courses to take. All-in-all, we find that the number of course schedules available for our student is: $6!*6*2*12*7=725760$.

    \item From 100 to 999, there are exactly 900 different numbers.If we are excluding all numbers that contain a 7, then we must first take out all numbers between 700-799, which is exactly 100 numbers. Then, since the number 7 comes up once in every ten numbers, we know that 1/10 if the remaining numbers contain 7's, so the final answer is 800-80=720.

    \item Since there are 10 questions on the exam and there are three different responses for each question (true, false, or blank), then the total number of ways the exam can be answered is $3^10 = 59049$.

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\\
\noindent {\bf Problem 2:}
\\ With ten students available for two positions, there will be exactly 90 different sets, since there will be 10 people available for the first position and 9 people available for the second. The same logic can be used for the next part of the problem, since there will be 10 people available for the presidency position, then 9 people available for the first executive slot and then 8 people for the second executive slot. Therefore, there are 720 different ways to arrange an executive board with a president.



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