14 quizzes are available for ECE2191 students, but students need to (and are only allowed to take) take any 10 out of all quizzes.  The quizzes are ranked in order of importance, so that the first quiz is the most important of all. If you are a student enrolled in this unit:How many choices of 10 selection of quizzes are available to you?  AnswerHow many are available if you must take the first two quizzes?   AnswerHow many are available if you must take the first quiz or second quiz but not both?    AnswerHow many if you must take exactly 3 of the first 5 tests?  AnswerHow many if you must take at least 3 out the first 5 tests? Answer

1. How many choices of 10 selection of quizzes are available to you?

This is a combination problem. The formula for combinations is C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

So, C(14, 10) = 14! / [10!(14-10)!] = 1001 choices.

2. How many are available if you must take the first two quizzes?

If you must take the first two quizzes, then you have 8 more quizzes to choose from the remaining 12. So, C(12, 8) = 12! / [8!(12-8)!] = 495 choices.

3. How many are available if you must take the first quiz or second quiz but not both?

If you must take either the first or the second quiz, then you have 9 more quizzes to choose from the remaining 13. So, C(13, 9) = 13! / [9!(13-9)!] = 715 choices.

4. How many if you must take exactly 3 of the first 5 tests?

If you must take exactly 3 of the first 5 tests, then you have 7 more quizzes to choose from the remaining 9. So, C(5, 3) * C(9, 7) = 10 * 36 = 360 choices.

5. How many if you must take at least 3 out the first 5 tests?

If you must take at least 3 of the first 5 tests, then you can take 3, 4, or 5 of the first 5 tests. So, C(5, 3) * C(9, 7) + C(5, 4) * C(9, 6) + C(5, 5) * C(9, 5) = 10 * 36 + 5 * 84 + 1 * 126 = 360 + 420 + 126 = 906 choices.