Question:
A student has to answer 10 out of 13 questions in an examination choosing atleast 3 from the 5 particular questions. The number of choices available to the student is
A student has to answer 10 out of 13 questions in an examination choosing atleast 3 from the 5 particular questions. The number of choices available to the student is
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When computing combinations like \( \binom{8}{6} \), always use the symmetry property \( \binom{n}{r} = \binom{n}{n-r} \). Thus, \( \binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2} = 28 \), which saves valuable time!
Updated On: Jun 7, 2026
- \( 196 \)
- \( 276 \)
- \( 326 \)
- \( 156 \)
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The Correct Option is B
Solution and Explanation
Concept:
The total number of questions is 13, divided into two distinct groups:
• Group I: 5 particular questions.
• Group II: Remaining questions \( = 13 - 5 = 8 \) questions.
The student needs to select a total of 10 questions such that at least 3 questions are chosen from Group I. We can break this into mutually exclusive cases based on the number of questions selected from Group I.
Step 1: Analyzing all valid selection cases.
Since at least 3 questions must be chosen from Group I (out of 5 available) and the total must equal 10:
• Case 1: 3 questions from Group I and 7 questions from Group II. \[ \text{Ways} = \binom{5}{3} \times \binom{8}{7} \]
• Case 2: 4 questions from Group I and 6 questions from Group II. \[ \text{Ways} = \binom{5}{4} \times \binom{8}{6} \]
• Case 3: 5 questions from Group I and 5 questions from Group II. \[ \text{Ways} = \binom{5}{5} \times \binom{8}{5} \]
Step 2: Calculating the combinations for each case.
• Case 1: \( \binom{5}{3} \times \binom{8}{7} = 10 \times 8 = 80 \)
• Case 2: \( \binom{5}{4} \times \binom{8}{6} = 5 \times 28 = 140 \)
• Case 3: \( \binom{5}{5} \times \binom{8}{5} = 1 \times 56 = 56 \)
Step 3: Summing up the total choices.
\[ \text{Total Choices} = 80 + 140 + 56 = 276 \]
• Group I: 5 particular questions.
• Group II: Remaining questions \( = 13 - 5 = 8 \) questions.
The student needs to select a total of 10 questions such that at least 3 questions are chosen from Group I. We can break this into mutually exclusive cases based on the number of questions selected from Group I.
Step 1: Analyzing all valid selection cases.
Since at least 3 questions must be chosen from Group I (out of 5 available) and the total must equal 10:
• Case 1: 3 questions from Group I and 7 questions from Group II. \[ \text{Ways} = \binom{5}{3} \times \binom{8}{7} \]
• Case 2: 4 questions from Group I and 6 questions from Group II. \[ \text{Ways} = \binom{5}{4} \times \binom{8}{6} \]
• Case 3: 5 questions from Group I and 5 questions from Group II. \[ \text{Ways} = \binom{5}{5} \times \binom{8}{5} \]
Step 2: Calculating the combinations for each case.
• Case 1: \( \binom{5}{3} \times \binom{8}{7} = 10 \times 8 = 80 \)
• Case 2: \( \binom{5}{4} \times \binom{8}{6} = 5 \times 28 = 140 \)
• Case 3: \( \binom{5}{5} \times \binom{8}{5} = 1 \times 56 = 56 \)
Step 3: Summing up the total choices.
\[ \text{Total Choices} = 80 + 140 + 56 = 276 \]
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