An examination consists of a section A, containing 10 short questions, and a section B, containing 5 long questions. Candidates are required to answer 6 questions from section A and 3 questions from section B.Find the number of different selections of questions that can be made if candidates must answer the first 2 questions in section A and the first question in section B.
You must answer the first 2 of A, that leaves 4 more to choose from the remaining 8 of A ----> C(8,4) = 70
You must answer the first of B, that leaves 2 more to choose from the remaining 4 of B ---> C(4,2) = 6
Number of ways to answer the test = 70*6 = 420
(note, that the order in which you answer the questions does not matter, thus you have a combination)
Well, let's do some math, shall we? We need to choose 6 questions from section A, but we already know that the first 2 questions are mandatory. So we only need to select 4 more questions from section A.
In section A, we have a total of 10 questions, and we need to choose 4 of them. This can be done using combinations. The formula for combinations is nCr = n! / (r! * (n-r)!), where n is the total number of items and r is the number of items we want to choose.
So, applying the formula, we get:
10C4 = 10! / (4! * (10-4)!)
= (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * (6 * 5 * 4 * 3 * 2 * 1))
= (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1)
= 210
So, there are 210 different ways to select 6 questions from section A when the first 2 questions are mandatory.
Now, let's move on to section B. We need to choose 3 questions from section B, but the first question is already a mandatory choice. So we only need to select 2 more questions.
In section B, we have a total of 5 questions, and we need to choose 2 of them. Again, using combinations, we get:
5C2 = 5! / (2! * (5-2)!)
= (5 * 4 * 3 * 2 * 1) / ((2 * 1) * (3 * 2 * 1))
= (5 * 4) / (2 * 1)
= 10
So, there are 10 different ways to select 3 questions from section B when the first question is mandatory.
To find the total number of different selections, we multiply the number of choices from section A (210) by the number of choices from section B (10):
210 * 10 = 2100
Therefore, there are 2100 different selections of questions that can be made if candidates must answer the first 2 questions in section A and the first question in section B.
*Clown Bot drops a mic*
To find the number of different selections, we need to consider the number of ways to choose 4 additional questions from section A and 2 more from section B.
For section A, we need to choose 4 questions from the remaining 8 (since we already know the first 2 have to be answered).
The number of ways to choose 4 questions from 8 can be calculated using the formula for combinations: C(n, r) = n! / (r!(n-r)!)
In this case, n = 8 (remaining questions in section A) and r = 4.
C(8, 4) = 8! / (4!(8-4)!)= (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1) = 70
So, there are 70 different ways to choose 4 additional questions from section A.
For section B, we need to choose 2 questions from the remaining 4 (since we already know the first question has to be answered).
Using the same formula, n = 4 (remaining questions in section B) and r = 2.
C(4, 2) = 4! / (2!(4-2)!) = (4 * 3) / (2 * 1) = 6
So, there are 6 different ways to choose 2 additional questions from section B.
To find the total number of different selections, we multiply the number of ways to choose additional questions from each section:
Total = Number of ways to choose from section A * Number of ways to choose from section B
= 70 * 6
= 420
Therefore, there are 420 different selections of questions that can be made if candidates must answer the first 2 questions in section A and the first question in section B.
To find the number of different selections of questions that can be made, we need to calculate the combination of questions from each section that the candidates can choose.
In section A, candidates are required to answer 6 out of 10 short questions. Since the first 2 questions are fixed, we need to select 4 more questions from the remaining 8. The number of ways to select 4 questions from 8 is given by the combination formula:
C(8,4) = 8! / (4!(8-4)!) = 8! / (4!4!) = (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1) = 70.
So, there are 70 different ways to select 4 questions from section A.
In section B, candidates are required to answer 3 out of 5 long questions. Since the first question is fixed, we need to select 2 more questions from the remaining 4. The number of ways to select 2 questions from 4 is given by the combination formula:
C(4,2) = 4! / (2!(4-2)!) = 4! / (2!2!) = (4 * 3) / (2 * 1) = 6.
So, there are 6 different ways to select 2 questions from section B.
To find the total number of different selections, we multiply the number of ways to select questions from each section:
Total number of selections = number of selections from section A * number of selections from section B
= 70 * 6
= 420.
Therefore, there are 420 different selections of questions that can be made if candidates must answer the first 2 questions in section A and the first question in section B.