Question

A question paper has 3 parts A, B, C. Part A contains 7 questions, part B contains 5 questions and Part C contains 3 questions. If a candidate is allowed to answer not more than 4 questions from part A; not more than 3 questions from part B and not more than 2 questions from part C, then the number of ways in which a candidate can answer exactly 7 questions is:

• A. 4655
• B. 4025
• C. 3675
• D. 2625 \bigskip

Hint:

For problems with restrictions, divide the problem into cases based on the number of questions from each part, then calculate the number of ways for each case.

Answer 1

The Correct Option is A

Step 1: The question paper consists of three sections: A, B, and C. We are tasked with determining the number of ways a candidate can answer exactly 7 questions, subject to the following conditions: - No more than 4 questions can be answered from part A. - No more than 3 questions can be answered from part B. - No more than 2 questions can be answered from part C. Let \( x_A \), \( x_B \), and \( x_C \) represent the number of questions answered from parts A, B, and C, respectively. We know: \[ x_A + x_B + x_C = 7. \] Additionally, the constraints are: \[ 0 \leq x_A \leq 4, \quad 0 \leq x_B \leq 3, \quad 0 \leq x_C \leq 2. \] 

Step 2: We now find the valid combinations for \( x_A \), \( x_B \), and \( x_C \) that satisfy the given constraints. The possible combinations where the sum of the questions equals 7 are: - \( x_A = 4, x_B = 3, x_C = 0 \) - \( x_A = 4, x_B = 2, x_C = 1 \) - \( x_A = 4, x_B = 1, x_C = 2 \) - \( x_A = 3, x_B = 3, x_C = 1 \) - \( x_A = 3, x_B = 2, x_C = 2 \) - \( x_A = 2, x_B = 3, x_C = 2 \) 

Step 3: For each valid combination, we calculate the number of ways to select the questions from each part: - For \( x_A = 4, x_B = 3, x_C = 0 \), the number of ways to choose the questions is: \[ \binom{7}{4} \times \binom{5}{3} \times \binom{3}{0} = 35 \times 10 \times 1 = 350. \] - For \( x_A = 4, x_B = 2, x_C = 1 \), the number of ways to choose the questions is: \[ \binom{7}{4} \times \binom{5}{2} \times \binom{3}{1} = 35 \times 10 \times 3 = 1050. \] - For \( x_A = 4, x_B = 1, x_C = 2 \), the number of ways to choose the questions is: \[ \binom{7}{4} \times \binom{5}{1} \times \binom{3}{2} = 35 \times 5 \times 3 = 525. \] - For \( x_A = 3, x_B = 3, x_C = 1 \), the number of ways to choose the questions is: \[ \binom{7}{3} \times \binom{5}{3} \times \binom{3}{1} = 35 \times 10 \times 3 = 1050. \] - For \( x_A = 3, x_B = 2, x_C = 2 \), the number of ways to choose the questions is: \[ \binom{7}{3} \times \binom{5}{2} \times \binom{3}{2} = 35 \times 10 \times 3 = 1050. \] - For \( x_A = 2, x_B = 3, x_C = 2 \), the number of ways to choose the questions is: \[ \binom{7}{2} \times \binom{5}{3} \times \binom{3}{2} = 21 \times 10 \times 3 = 630. \] 

Step 4: The total number of ways is obtained by summing the values calculated for each valid combination: \[ 350 + 1050 + 525 + 1050 + 1050 + 630 = 4655. \] Therefore, the total number of ways the candidate can answer exactly 7 questions is \( \boxed{4655} \). \bigskip