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Chapter 5: Problem 424

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choice available to him is (a) 140 (b) 196 (c) 180 (d) 346

Short Answer

Expert verified
The number of choices available to the student is \(( _5 C _4 * _8 C _6 ) + ( _5 C _5 * _8 C _5 ) = (5 * 28) + (1 * 56) = 196\). Hence, the correct answer is (b) 196.

Step by step solution

01

Case 1: The student answers exactly 4 out of the first 5 questions.

In this situation, the student will choose 4 out of the first 5 questions and the remaining 6 questions out of the last 8 questions. We can calculate this using combinations: Number of ways to choose 4 out of 5 questions: \( _5 C _4 \) Number of ways to choose 6 out of the last 8 questions: \( _8 C _6 \) Total number of ways in this case: \( _5 C _4 * _8 C _6 \)
02

Case 2: The student answers all 5 out of the first 5 questions.

In this situation, the student will choose all 5 out of the first 5 questions and the remaining 5 questions out of the last 8 questions. We can calculate this using combinations: Number of ways to choose 5 out of 5 questions: \( _5 C _5 \) Number of ways to choose 5 out of the last 8 questions: \( _8 C _5 \) Total number of ways in this case: \( _5 C _5 * _8 C _5 \)
03

Total number of choices available:

To find the total number of choices, we sum the number of ways in both cases: Total choices = (Number of ways in Case 1) + (Number of ways in Case 2) Total choices = \(( _5 C _4 * _8 C _6 ) + ( _5 C _5 * _8 C _5 ) \) Now, we will compute the value for each combination and then add them up to get the final answer: \( _5 C _4 = \frac{5!}{4!1!} = 5 \) \( _8 C _6 = \frac{8!}{6!2!} = 28 \) \( _5 C _5 = \frac{5!}{5!0!} = 1 \) \( _8 C _5 = \frac{8!}{5!3!} = 56 \) Total choices = (5 * 28) + (1 * 56) Total choices = 140 + 56 Total choices = 196 The correct answer is (b) 196. The student has 196 different choices on how to answer the examination questions.

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