A physics exam consists of 9 multiple choice questions and 6 open-ended problems in which all work must be shown.If an examinee must answer 6 of the multiple- choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen?
9C6 * 6C3
To find the number of ways the questions and problems can be chosen, we need to calculate the combinations.
For the multiple-choice questions, the examinee must answer 6 out of 9. The number of ways to choose 6 out of 9 is given by the formula C(n, r) = n! / (r!(n-r)!), where n is the total number of choices and r is the number of choices to be made.
Using this formula, we have C(9, 6) = 9! / (6!(9-6)!) = 84.
For the open-ended problems, the examinee must answer 3 out of 6. Again, using the combination formula, we have C(6, 3) = 6! / (3!(6-3)!) = 20.
To find the total number of ways the questions and problems can be chosen, we multiply these two results together: 84 * 20 = 1680.
Therefore, there are 1680 ways the questions and problems can be chosen.
To solve this problem, we need to find the number of ways the multiple-choice questions and open-ended problems can be chosen separately, and then multiply those two numbers together to get the total number of ways.
For the multiple-choice questions, the examinee needs to answer 6 out of 9 questions. We can use the combination formula to calculate the number of ways to choose 6 questions from 9. The formula is:
C(n, r) = n! / (r! * (n - r)!)
where n is the total number of questions and r is the number of questions to be chosen. Plugging in the values:
C(9, 6) = 9! / (6! * (9 - 6)!)
= 9! / (6! * 3!)
= (9 * 8 * 7 * 6!) / (6! * (3 * 2 * 1))
= (9 * 8 * 7) / (3 * 2 * 1)
= 84
So, there are 84 ways to choose 6 multiple-choice questions.
For the open-ended problems, the examinee needs to answer 3 out of 6 problems. Using the same combination formula:
C(6, 3) = 6! / (3! * (6 - 3)!)
= 6! / (3! * 3!)
= (6 * 5 * 4 * 3!) / (3! * (3 * 2 * 1))
= (6 * 5 * 4) / (3 * 2 * 1)
= 20
So, there are 20 ways to choose 3 open-ended problems.
To find the total number of ways to choose both the multiple-choice questions and open-ended problems, we multiply the two numbers together:
Total ways = 84 * 20
= 1680
Therefore, there are 1680 ways to choose the questions and problems.