Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects, denoted by $\binom{n}{r}$ or $^nC_r$, is given by $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
• Diophantine Equations: Equations where we seek integer solutions.
• Scoring System: Understanding how correct and incorrect answers contribute to the total score.

Step-by-Step Solution

Step 1: Define Variables and Set up the Equation

Let $c$ be the number of correct answers, $w$ be the number of wrong answers, and $u$ be the number of unattempted questions. We know that the total number of questions is 5, so $c + w + u = 5$ The total score is given by $3c - 2w + 0u = 5$. We are looking for non-negative integer solutions for $c$, $w$, and $u$.

Step 2: Analyze Possible Values of c

Since $3c \le 5 + 2w$ and $c$ and $w$ are non-negative integers, we can analyze possible values for $c$.

• If $c = 0$, then $-2w = 5$, which has no non-negative integer solution for $w$.
• If $c = 1$, then $3(1) - 2w = 5$, which gives $-2w = 2$, so $w = -1$. This is not possible since $w$ must be non-negative.
• If $c = 2$, then $3(2) - 2w = 5$, which gives $6 - 2w = 5$, so $-2w = -1$, and $w = \frac{1}{2}$. This is not an integer, so it's not a valid solution.
• If $c = 3$, then $3(3) - 2w = 5$, which gives $9 - 2w = 5$, so $-2w = -4$, and $w = 2$. Then, $u = 5 - c - w = 5 - 3 - 2 = 0$. This is a valid solution: (c, w, u) = (3, 2, 0).
• If $c = 4$, then $3(4) - 2w = 5$, which gives $12 - 2w = 5$, so $-2w = -7$, and $w = \frac{7}{2}$. This is not an integer, so it's not a valid solution.
• If $c = 5$, then $3(5) - 2w = 5$, which gives $15 - 2w = 5$, so $-2w = -10$, and $w = 5$. Then $u = 5 - 5 - 5 = -5$. This is not possible since $u$ must be non-negative.

So, the only possible solution in non-negative integers is $c = 3$, $w = 2$, and $u = 0$.

Step 3: Calculate the Number of Ways to Achieve the Solution

We have 3 correct answers and 2 wrong answers. We need to choose which 3 questions are answered correctly and which 2 are answered incorrectly. Since the order in which the questions are answered does not matter, we use combinations.

First, we choose 3 questions to be answered correctly out of 5 questions: $\binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2} = 10$.

Then, from the remaining 2 questions, we choose 2 questions to be answered incorrectly: $\binom{2}{2} = \frac{2!}{2!0!} = 1$.

For each correct answer, there is only 1 way to answer it correctly (since only one choice is correct). For each wrong answer, there are 2 ways to answer it incorrectly (since 2 choices are incorrect). So, for 2 wrong answers, there are $2 \times 2 = 2^2 = 4$ ways.

Therefore, the number of ways to get 3 correct and 2 wrong answers is: $\binom{5}{3} \times \binom{2}{2} \times 1^3 \times 2^2 = 10 \times 1 \times 1 \times 4 = 40$.

Step 4: Re-evaluate the Approach and Identify Error

The above approach is incorrect. The problem asks for the number of ways a student gets a score of 5, not the probability. We found the correct values of $c, w, u$, but miscalculated the number of ways to achieve the score. We need to choose 3 questions to be correct ($\binom{5}{3}$), and then choose 2 questions to be incorrect from the remaining 2 questions ($\binom{2}{2}$). The number of ways to answer the 3 correct questions is 1 for each, and the number of ways to answer the 2 incorrect questions is 2 for each.

So, the number of ways is $\binom{5}{3} \cdot 2^2 = 10 \cdot 4 = 40$. This is still not correct.

Let's carefully consider the situation: We need 3 correct answers and 2 wrong answers. The number of ways to choose the 3 correct answers from 5 questions is $\binom{5}{3} = 10$. Then, from the remaining 2 questions, we choose 2 to be wrong answers. For each wrong answer, there are 2 choices. So we have $\binom{5}{3} \times 2^2 = 10 \times 4 = 40$. However, we are given that the correct answer is 3. This implies that we have made a mistake.

Step 5: Correct Solution

We have $3c - 2w = 5$ and $c + w + u = 5$. We found $c=3, w=2, u=0$ as the only valid integer solution. So, a student needs to answer exactly 3 questions correctly and 2 questions wrongly. We need to select 3 questions out of 5 to be correct. This can be done in $\binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2} = 10$ ways. Once these 3 questions are selected, they must be answered correctly, and since each question has only one correct answer, there is only 1 way to answer each of them correctly. Now, we have 2 questions remaining. These must be answered wrongly. For each question, there are 2 incorrect options. Thus, each of these 2 questions can be answered wrongly in 2 ways. Hence, the number of ways to answer these 2 questions wrongly is $2 \times 2 = 4$. Therefore, the total number of ways to get exactly 5 marks is $10 \times 1 \times 1 \times 1 \times 2 \times 2 = 10 \times 4 = 40$.

However, the correct answer is given as 3. Let's re-think this again.

If we have 3 correct, the score is 9. To get to 5, we need to subtract 4. We can do this with 2 wrong answers. $c+w+u = 5$ $3c-2w = 5$

If $c=3$, then $w=2$ and $u=0$. Number of ways = $\binom{5}{3} \binom{2}{2} (1)^3 (2)^2 = 10 \times 1 \times 1 \times 4 = 40$.

If $c=1$, then $3-2w=5$ so $-2w=2$ and $w=-1$, which is not possible. If $c=2$, then $6-2w=5$ so $2w=1$ and $w=\frac{1}{2}$, which is not possible.

Let's try the cases: Case 1: 3 correct, 2 wrong, 0 unattempted. Choose 3 correct in $\binom{5}{3} = 10$ ways. Then for the other 2 questions, we must answer them wrong. Each of these 2 can be answered wrongly in 2 ways. So $2 \times 2 = 4$. Total is $10 \times 4 = 40$.

There MUST be an error in the question or answer. I will try to find out by working backwards.

If answer is 3. Case 1: 3 correct, 2 wrong. This leads to $10 \times 2 \times 2 = 40$.

There must be a constraint that the student attempts all the questions. Then $u=0$. So $c+w=5$ and $3c-2w=5$. Then $3c - 2(5-c)=5$ $3c-10+2c=5$ $5c=15$ $c=3$ $w=2$ Then we select 3 out of 5 questions correctly.

Let's assume that the student MUST attempt all questions. Then $c+w = 5$, so $w = 5 - c$. $3c - 2(5-c) = 5$ $3c - 10 + 2c = 5$ $5c = 15$ $c = 3$. Then $w = 2$. So we must have 3 correct and 2 wrong. The number of ways to choose 3 questions to be correct is $\binom{5}{3} = 10$. For each of the other questions, there are 2 wrong choices. So $2 \times 2 = 4$. So $10 \cdot 4 = 40$.

However, the correct answer is 3. This is a contradiction. There must be an error in the question or the provided solution.

Let us assume that at most one question can be incorrect.

If $c = 5, w = 0, u = 0$, then score = $3(5) = 15$. If $c = 4, w = 1, u = 0$, then score = $3(4) - 2(1) = 10$. If $c = 3, w = 0, u = 2$, then score = $3(3) = 9$. If $c = 3, w = 1, u = 1$, then score = $3(3) - 2(1) = 7$.

Assuming the intended answer is 40, the solution is:

Step 1: Define Variables and Set up the Equation

Let $c$ be the number of correct answers, $w$ be the number of wrong answers, and $u$ be the number of unattempted questions. We know that the total number of questions is 5, so $c + w + u = 5$ The total score is given by $3c - 2w + 0u = 5$. We are looking for non-negative integer solutions for $c$, $w$, and $u$.

Step 2: Find possible values of c, w, u

We need to find non-negative integer solutions for $c, w, u$. $3c = 5 + 2w$ If $c=3$, $9 = 5 + 2w$ so $2w=4$ and $w=2$. Then $3+2+u=5$ so $u=0$. So $c=3, w=2, u=0$ is the only solution.

Step 3: Calculate the number of ways to achieve the solution We have 3 correct answers and 2 wrong answers. We need to choose which 3 questions are answered correctly and which 2 are answered incorrectly. First, we choose 3 questions to be answered correctly out of 5 questions: $\binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2} = 10$.

Then, from the remaining 2 questions, we choose 2 questions to be answered incorrectly: $\binom{2}{2} = \frac{2!}{2!0!} = 1$.

For each correct answer, there is only 1 way to answer it correctly (since only one choice is correct). For each wrong answer, there are 2 ways to answer it incorrectly (since 2 choices are incorrect). So, for 2 wrong answers, there are $2 \times 2 = 2^2 = 4$ ways.

Therefore, the number of ways to get 3 correct and 2 wrong answers is: $\binom{5}{3} \times 2^2 = 10 \times 4 = 40$.

Common Mistakes & Tips

• Carefully analyze the constraints of the problem.
• Ensure that all variables are non-negative integers.
• Be careful when calculating the number of ways to answer questions incorrectly.

Summary

We analyzed the problem by setting up equations based on the given scoring system and the total number of questions. We found that the only possible solution in non-negative integers is 3 correct answers and 2 wrong answers. We then calculated the number of ways to achieve this solution, which is 40. The provided "Correct Answer" of 3 seems incorrect.

Final Answer

The final answer is \boxed{40}.