Permutations and CombinationsMCQMTP Sep 24 Series IQuestion 1752 of 251

$\displaystyle ^{n}C_1 + ^{n}C_2 + ^{n}C_3 + ............$

Options

\displaystyle 2^{n}-1

\displaystyle 2^{n}-2

\displaystyle 2^{n}+1

DNone of these

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Correct Answer

✅ Option a — $\displaystyle 2^{n}-1$

All Options:

A $\displaystyle 2^{n}-1$
B $\displaystyle 2^{n}-2$
C $\displaystyle 2^{n}+1$
DNone of these

Detailed Solution & Explanation

The question asks for the sum of the series of binomial coefficients starting from

\displaystyle ^{n}C_{1}

S = ^{n}C_1 + ^{n}C_2 + ^{n}C_3 + \dots + ^{n}C_n

We know from the Binomial Theorem expansion of

\displaystyle (1 + x)^n

(1 + x)^n = \sum_{r=0}^{n} {^{n}C_r x^r} = ^{n}C_0 + ^{n}C_1 x + ^{n}C_2 x^2 + \dots + ^{n}C_n x^n

Substituting

\displaystyle x = 1

into this identity:

(1 + 1)^n = ^{n}C_0 + ^{n}C_1 + ^{n}C_2 + \dots + ^{n}C_n

\implies 2^n = ^{n}C_0 + \left(^{n}C_1 + ^{n}C_2 + \dots + ^{n}C_n\right)

Since

\displaystyle ^{n}C_0 = \frac{n!}{0! \times n!} = 1

2^n = 1 + \left(^{n}C_1 + ^{n}C_2 + \dots + ^{n}C_n\right)

\implies ^{n}C_1 + ^{n}C_2 + \dots + ^{n}C_n = 2^n - 1

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

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nC1+nC2+nC3+............\displaystyle ^{n}C_1 + ^{n}C_2 + ^{n}C_3 + ............nC1​+nC2​+nC3​+............