A student s to answer 10 out of 13 questions in an exam. How many choices he has if he must answer at least 3 of the first 5 questions.
Assuming that he can answer them in any order and that he is not restricted to anwering only 3 of the first 5, that is, he could answer all of the first five.
step one: number of ways to answer the obligatory questions = C(5,3) = 10
So now he has to answer 7 more of any of the remaining 10
= C(10,7) = 120
number of ways = 12(120) = 1200
a women student is to answer b10 out of 13 questions.find the number of her choices where she must answer
a,the frist two question
b,exactly 3 out of the frist 5 question
c,the first or second question but not both;
at least3 of the frist 5 question
To calculate the number of choices the student has in this scenario, we need to consider the two possible cases:
Case 1: The student chooses to answer exactly 3 of the first 5 questions.
Case 2: The student chooses to answer more than 3 of the first 5 questions.
Step 1: Calculate the number of choices for each case separately.
Case 1:
In this case, the student must choose 3 questions out of the first 5 and any 7 questions out of the remaining 8 questions. To calculate the number of choices, we can use the combination formula, also known as "nCr":
Number of choices in Case 1 = C(5,3) * C(8,7)
= (5! / (3!(5-3)!) ) * (8! / (7!(8-7)!) )
= (5! / (3! × 2!) ) * (8! / (7! × 1!) )
= (5 × 4 × 3! / (3! × 2!) ) * (8!/7!×1! )
= (5 × 4 / 2) * (8 × 7)
= 10 * 56
= 560
Case 2:
In this case, the student can choose any number of questions (greater than 3) out of the first 5 and any remaining number of questions from the remaining 8 questions. Since the student can choose any number greater than 3, we can sum up the combinations for choosing 4, 5, and all 6 questions from the first 5:
Number of choices in Case 2 = C(5,4) * C(8,6) + C(5,5) * C(8,5) + C(5,6) * C(8,4)
= (5! / (4!(5-4)!) ) * (8! / (6!(8-6)!) ) + (5! / (5!(5-5)!) ) * (8! / (5!(8-5)!) ) + (5! / (6!(5-6)!) ) * (8! / (4!(8-4)!) )
= (5! / (4! × 1!) ) * (8! / (6! × 2!) ) +(5! / (5! × 0!) ) * (8! / (5! × 3!) ) +(5! / (6! × (-1!) ) ) * (8! / (4! × 4!) )
= (5 × 8) + (1 × 56) + (0 × 70)
= 40 + 56 + 0
= 96
Step 2: Add the number of choices in both cases together to get the total number of choices:
Total number of choices = Number of choices in Case 1 + Number of choices in Case 2
= 560 + 96
= 656
Therefore, the student has a total of 656 choices if they need to answer at least 3 of the first 5 questions.
To find the number of choices the student has, we need to consider two scenarios:
1. The student answers exactly 3, 4, or 5 questions from the first 5 questions.
2. The student answers more than 5 questions from the first 5 questions.
Scenario 1: Answering 3, 4, or 5 questions from the first 5 questions.
For each case, we need to calculate the number of choices for the remaining questions (Q6 to Q13).
For the first 5 questions (Q1-Q5), the student has the following choices:
- To answer 3 questions: Choose 3 out of 5, which can be done using the combination formula: C(5, 3) = 5! / (3! * (5 - 3)!) = 10 choices.
- To answer 4 questions: Choose 4 out of 5, using the combination formula: C(5, 4) = 5! / (4! * (5 - 4)!) = 5 choices.
- To answer 5 questions: Choose all 5 questions, so there is only 1 choice.
For the remaining questions (Q6-Q13), the student will have to answer 10 minus the number of questions answered from the first 5. Thus, the number of choices for each case is:
- 3 questions from Q6-Q13: Choose 7 out of 8, using the combination formula: C(8, 7) = 8 choices.
- 4 questions from Q6-Q13: Choose 6 out of 8, using the combination formula: C(8, 6) = 28 choices.
- 5 questions from Q6-Q13: Choose 5 out of 8, using the combination formula: C(8, 5) = 56 choices.
Scenario 2: Answering more than 5 questions from the first 5 questions.
In this case, the student can choose 6, 7, 8, or 9 questions from the first 5.
For each case, we calculate the number of choices for the remaining questions (Q6 to Q13).
- 6 questions from Q1-Q5: Choose 6 out of 5, which is not possible, so there are 0 choices.
- 7 questions from Q1-Q5: Choose 7 out of 5, again not possible, so 0 choices.
- 8 questions from Q1-Q5: Choose 8 out of 5, also not possible, so 0 choices.
- 9 questions from Q1-Q5: Choose 9 out of 5, still not possible, so 0 choices.
Summing up the choices for each scenario:
- Scenario 1: 10 choices for answering 3 questions from Q1-Q5, 5 choices for answering 4 questions, and 1 choice for answering all 5 questions. For each case, there are 8 choices for Q6-Q13.
- Scenario 2: Not possible to answer more than 5 questions, so 0 choices.
Adding up the choices for each scenario:
(10 + 5 + 1) * 8 + 0 = 16 * 8 = 128 choices.
Therefore, the student has 128 choices if he must answer at least 3 of the first 5 questions.