Question 38 of 42intermediate🔧 ApplyNumerical4 marks

In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Correct Answer

Total number of questions = 12. Part I contains 5 questions and Part II contains 7 questions. The student must attempt 8 questions selecting at least 3 from each part.

The selection can be made in the following three ways: (i) 3 questions from Part I and 5 questions from Part II. (ii) 4 questions from Part I and 4 questions from Part II. (iii) 5 questions from Part I and 3 questions from Part II.

Case I: Selecting 3 questions from Part I and 5 questions from Part II. Number of ways = 5C3×7C5^5C_3 \times ^7C_5

=10×21=210= 10 \times 21 = 210

Case II: Selecting 4 questions from Part I and 4 questions from Part II. Number of ways = 5C4×7C4^5C_4 \times ^7C_4

=5×35=175= 5 \times 35 = 175

Case III: Selecting 5 questions from Part I and 3 questions from Part II. Number of ways = 5C5×7C3^5C_5 \times ^7C_3

=1×35=35= 1 \times 35 = 35

Total number of ways to select questions = 210+175+35=420210 + 175 + 35 = 420.

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 7 | (Page 24)
For More Understanding

Explanation

This problem is solved using the concept of combinations (nCr^nC_r) as the order of selection does not matter. Similar to the committee selection examples provided in the context (like Example 3 and the Miscellaneous Example), we break the problem into distinct cases based on the 'at least' constraint. Since the student must attempt 8 questions total with at least 3 from each part, the possible distributions of questions between Part I and Part II are (3,5), (4,4), and (5,3). We calculate the ways for each case and add them together.

Solution Steps

  1. Step 1: Identify the distribution of questions. Part I has 5 questions and Part II has 7. We need to select 8 questions in total with at least 3 from each part.

  2. Step 2: List the possible cases. The valid combinations are: (3 from Part I, 5 from Part II), (4 from Part I, 4 from Part II), and (5 from Part I, 3 from Part II).

  3. Step 3: Calculate ways for Case I. 5C3×7C5=10×21=210^5C_3 \times ^7C_5 = 10 \times 21 = 210.

  4. Step 4: Calculate ways for Case II. 5C4×7C4=5×35=175^5C_4 \times ^7C_4 = 5 \times 35 = 175.

  5. Step 5: Calculate ways for Case III. 5C5×7C3=1×35=35^5C_5 \times ^7C_3 = 1 \times 35 = 35.

  6. Step 6: Sum the results. 210+175+35=420210 + 175 + 35 = 420.

PreviousNext