Permutations and CombinationsMCQMTP May 19 IIQuestion 1724 of 251
All Questions

A man has 5 friends. In how many ways can be invite one or more of his friends to dinner?

Options

A30
B31
C32
D10
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — 31

All Options:

  • A30
  • B31
  • C32
  • D10

Ad

Detailed Solution & Explanation

The man has 5\displaystyle 5 friends. He wants to invite **one or more** of his friends to dinner.

For each of the 5\displaystyle 5 friends, there are 2\displaystyle 2 possible choices:
1. The friend is **invited**
2. The friend is **not invited**

By the multiplication principle, the total number of ways to make choices for all 5\displaystyle 5 friends is:
Total Ways=25=32\text{Total Ways} = 2^5 = 32
However, these 32\displaystyle 32 ways include the scenario where **none** of the friends are invited (each friend is "not invited"). Since the problem states that he must invite **one or more** (i.e., at least 1\displaystyle 1) of his friends, we must subtract the 1\displaystyle 1 way in which nobody is invited:
Ways to invite one or more=25−1=32−1=31\text{Ways to invite one or more} = 2^5 - 1 = 32 - 1 = 31
Alternatively, this can be calculated as the sum of choosing exactly 1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4, or 5\displaystyle 5 friends:
Ways=5C1+5C2+5C3+5C4+5C5\text{Ways} = ^{5}C_{1} + ^{5}C_{2} + ^{5}C_{3} + ^{5}C_{4} + ^{5}C_{5}
Ways=5+10+10+5+1=31\text{Ways} = 5 + 10 + 10 + 5 + 1 = 31
**Discrepancy Note:**
The rigorous mathematical derivation yields 31\displaystyle 31 ways, which corresponds to **Option B**. The textbook answer key has a typographical error, indicating **Option C** (32\displaystyle 32) as the correct answer. This error occurs because the author neglected to subtract the case where no friends are invited.

Hence, **Option B** 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