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 5 of the multiple-choice questions and 2 of the open-ended problems, in how many ways can the questions and problems be chosen?
ways to choose 5 of the 6 multiple choices
= C(9,5) = 126
number of ways to choose 2 of the 6 essay questions
= C(6,2) = 15
number of ways to answer the test
= 126x15 or 1890 ways
1890
Well, let's see. If there are 9 multiple-choice questions and you need to answer 5, that means you have more options than the latest iPhone has features. But wait, there's more! Then we have 6 open-ended problems and you only need to answer 2. It's like trying to choose your favorite ice cream flavor when you're in an ice cream parlor with 31 different flavors.
To find out how many ways you can choose the questions and problems, we can use some mathematical magic. First, we need to find the number of ways to choose 5 from 9 multiple-choice questions, which can be calculated using combinations. This can be represented by "9 choose 5", which equals 126.
Similarly, we need to find the number of ways to choose 2 from 6 open-ended problems, or "6 choose 2", which equals 15.
To calculate the total number of ways you can choose both the questions and problems, we simply multiply these values together: 126 * 15 = 1890.
So, there are 1890 different ways in which you can choose the questions and problems for this physics exam. That's enough options to make your head spin faster than a particle accelerator!
To calculate the total number of ways the questions and problems can be chosen, we need to multiply the number of ways the multiple-choice questions can be chosen by the number of ways the open-ended problems can be chosen.
The number of ways to choose 5 multiple-choice questions out of 9 is given by the combination formula:
C(n, r) = n! / (r!(n-r)!)
Substituting n = 9 and r = 5:
C(9, 5) = 9! / (5!(9-5)!)
= 9! / (5!4!)
Simplifying the expression:
9! = 9 × 8 × 7 × 6 × 5!
C(9, 5) = (9 × 8 × 7 × 6 × 5!) / (5! × 4!)
The 5! cancels out from both the numerator and denominator:
C(9, 5) = 9 × 8 × 7 × 6 / 4!
= 9 × 8 × 7 × 6 / (4 × 3 × 2 × 1)
= 9 × 8 × 7 × 6 / 24
Simplifying further:
C(9, 5) = 9 × 2 × 7 × 6 / 1
= 9 × 14 × 6
= 756
Therefore, there are 756 ways to choose 5 multiple-choice questions out of 9.
Now let's calculate the number of ways to choose 2 open-ended problems out of 6 using the combination formula again:
C(6, 2) = 6! / (2!(6-2)!)
Substituting n = 6 and r = 2:
C(6, 2) = 6! / (2! × 4!)
= (6 × 5 × 4!) / (2! × 4!)
The 4! cancels out from both the numerator and denominator:
C(6, 2) = 6 × 5 / (2 × 1)
= 30 / 2
= 15
Therefore, there are 15 ways to choose 2 open-ended problems out of 6.
To find the total number of ways the questions and problems can be chosen, we multiply these two results:
Total ways = 756 × 15
= 11340
So, the questions and problems can be chosen in 11,340 different ways.
To find the number of ways the questions and problems can be chosen, we need to consider both the multiple-choice questions and the open-ended problems separately.
For the multiple-choice questions, the examinee must answer 5 out of 9. This can be done in C(9, 5) ways, which represents choosing 5 questions out of a total of 9. The combination formula C(n, r) can be calculated as:
C(n, r) = n! / (r! * (n - r)!)
Applying this formula to the given situation, we have:
C(9, 5) = 9! / (5! * (9 - 5)!)
= (9 * 8 * 7 * 6 * 5!) / (5! * 4 * 3 * 2 * 1)
= (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1)
= 3024 / 24
= 126
Therefore, there are 126 ways to choose 5 multiple-choice questions.
For the open-ended problems, the examinee must answer 2 out of 6. This can be done in C(6, 2) ways, using the same combination formula:
C(6, 2) = 6! / (2! * (6 - 2)!)
= (6 * 5 * 4 * 3 * 2!) / (2! * 4 * 3 * 2 * 1)
= (6 * 5) / (2 * 1)
= 15
Therefore, there are 15 ways to choose 2 open-ended problems.
To find the total number of ways the questions and problems can be chosen together, you need to multiply the number of ways to choose the multiple-choice questions by the number of ways to choose the open-ended problems:
Total number of ways = Number of ways to choose multiple-choice questions * Number of ways to choose open-ended problems
= 126 * 15
= 1890
Therefore, there are 1890 ways to choose the questions and problems in the physics exam.