Permutations and CombinationsMCQMTP March 22Question 1732 of 251
All Questions

A boy has 3 library tickets and 8 books of his interest in the library. Out of these 8, he does not want to borrow mathematics part II unless mathematics part-I is also borrowed. In how many ways can he choose the three books to be borrowed?

Options

A41
B51
C61
D71
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 41

All Options:

  • A41
  • B51
  • C61
  • D71

Ad

Detailed Solution & Explanation

Let the 8\displaystyle 8 books be represented as follows:
- M1\displaystyle M_1: Mathematics Part-I
- M2\displaystyle M_2: Mathematics Part-II
- The remaining 6\displaystyle 6 other books of interest.

The boy needs to select exactly 3\displaystyle 3 books using his 3\displaystyle 3 library tickets.
The condition is: He does not want to borrow M2\displaystyle M_2 unless M1\displaystyle M_1 is also borrowed. This implies two mutually exclusive and exhaustive cases:

**Case 1: The boy borrows Mathematics Part-II (M2\displaystyle M_2)**
According to the condition, if he borrows M2\displaystyle M_2, he **must** also borrow M1\displaystyle M_1.
Thus, 2\displaystyle 2 of the 3\displaystyle 3 book selections are already fixed: {M1,M2}\displaystyle \{M_1, M_2\}. He must select the remaining 3−2=1\displaystyle 3 - 2 = 1 book from the other 6\displaystyle 6 books.
Ways1=6C1=6\text{Ways}_1 = ^{6}C_{1} = 6

**Case 2: The boy does not borrow Mathematics Part-II (M2\displaystyle M_2)**
If he does not borrow M2\displaystyle M_2, he can select any 3\displaystyle 3 books from the remaining 7\displaystyle 7 books (which includes M1\displaystyle M_1 and the other 6\displaystyle 6 books).
Ways2=7C3=7×6×53×2×1=35\text{Ways}_2 = ^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

**Total number of ways to select the 3 books:**
Total Ways=Ways1+Ways2\text{Total Ways} = \text{Ways}_1 + \text{Ways}_2
Total Ways=6+35=41\text{Total Ways} = 6 + 35 = 41

Hence, **Option A** is the correct answer.

About This Chapter: Permutations and Combinations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Factorials, Permutations, Combinations

This chapter deals with the fundamental principles of counting. It covers factorials, circular permutations, restricted permutations, combinations, and the differences between selecting items versus arranging them.

View Official ICAI Syllabus

Exam Strategy Tip

The most common mistake is confusing 'P' (Arrangement) with 'C' (Selection). If order matters (like opening a lock), use P. If order doesn't matter (like choosing a team), use C.

Related Comparison Tables

Permutations vs Combinations: The Quantitative Aptitude Guide

More Questions from Permutations and Combinations

The value of N\displaystyle N in N!+17!=18!\displaystyle N! + \frac{1}{7!} = \frac{1}{8!} is

A person can go from place 'A' to 'B' by 11 different modes of transport but is allowed to return to 'A' by any mode other than the one earlier. The number of different ways in which the entire journey can be completed is:

If a man travels from place A to B in 10 ways then by how many ways can he come back by another train?

If n!10(n−1)!=(n−1)!(n−3)!\displaystyle \frac{n!}{10(n-1)!} = \frac{(n-1)!}{(n-3)!} find 'n'.

Which of the following is a correct statement.

nP3=2⋅nP2\displaystyle ^nP_3 = 2 \cdot ^nP_2. Find n\displaystyle n.

Ready to Master Permutations and Combinations?

Practice all 251 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free