There are 12 questions on an examination, and each student must answer 8 questions including at least 4 of the first 5 questions. How many different combinations of questions could a student choose to answer?
5C4x7C4=175
5C5x7C3=35
175+35=210
There are 210 different combinations.
To calculate the number of different combinations a student could choose to answer, we need to consider the conditions given.
The student must answer 8 questions in total, including at least 4 of the first 5 questions. Let's break it down into two cases:
Case 1: The student answers exactly 4 of the first 5 questions.
In this case, the student has to choose 4 questions out of the first 5 questions, and 4 questions out of the remaining 7 questions (12 total questions - 5 first questions = 7 remaining questions).
The number of ways to choose 4 questions out of 5 is given by the binomial coefficient "5 choose 4" or C(5,4), which is equal to 5.
The number of ways to choose 4 questions out of 7 is given by the binomial coefficient "7 choose 4" or C(7,4), which is equal to 35.
So, the number of combinations in this case is 5 * 35 = 175.
Case 2: The student answers more than 4 of the first 5 questions.
In this case, the student can choose 5, 6, 7, or all 8 questions from the first 5 questions. For each option, they can choose the remaining questions from the remaining 7 questions.
The number of ways to choose 5 questions out of 5 is given by the binomial coefficient "5 choose 5" or C(5,5), which is equal to 1.
The number of ways to choose 6 questions out of 5 is given by the binomial coefficient "5 choose 6" or C(5,6). Since choosing more elements than available is not possible, this binomial coefficient is zero.
Similarly, for choosing 7 questions out of 5 is C(5,7) and choosing 8 questions out of 5 is C(5,8), both of which are zero.
For the remaining 7 questions, the student needs to choose 3, 2, 1, or 0 questions. The respective binomial coefficients are C(7,3) = 35, C(7,2) = 21, C(7,1) = 7, and C(7,0) = 1.
Adding up all the possibilities in Case 2 gives us:
(1 * 35) + (0 * 21) + (0 * 7) + (0 * 1) = 35.
Finally, to get the total number of different combinations, we sum up the possibilities in both cases:
175 + 35 = 210.
Therefore, a student could choose from 210 different combinations of questions to answer.
To find out the number of different combinations of questions a student could choose to answer, we need to consider two cases:
Case 1: Choosing exactly 4 questions from the first 5 questions.
In this case, the student needs to choose 4 questions from the first 5, and 4 questions from the remaining 7 questions.
The combination formula could be used to calculate the number of ways to choose questions:
C(n, k) = n! / (k! * (n - k)!)
Using the combination formula, we can calculate the number of ways to choose 4 questions from the first 5:
C(5, 4) = 5! / (4! * (5 - 4)!) = 5
Similarly, we calculate the number of ways to choose 4 questions from the remaining 7:
C(7, 4) = 7! / (4! * (7 - 4)!) = 35
Case 2: Choosing more than 4 questions from the first 5 questions.
In this case, the student needs to choose 5 or more questions from the first 5, and the remaining questions from the remaining 7.
We can calculate the number of ways to choose 5 questions from the first 5:
C(5, 5) = 5! / (5! * (5 - 5)!) = 1
To find the number of ways to choose more than 4 questions from the first 5, we subtract the previous cases from the total number of combinations possible:
Total combinations = C(12, 8) = 12! / (8! * (12 - 8)!) = 495
Number of ways to choose exactly 4 questions from the first 5: 5 * 35 = 175
Number of ways to choose exactly 5 questions from the first 5: 1 * 35 = 35
Total combinations with more than 4 questions from the first 5: 495 - 175 - 35 = 285
Therefore, the total number of different combinations of questions a student could choose to answer is: 175 + 35 + 285 = 495.