Permutations and CombinationsMCQMTP May 18Question 1723 of 251
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nCr+nCr1+nCr2+...\displaystyle ^nC_r + ^nC_{r-1} + ^nC_{r-2} + ...

Options

A2n1\displaystyle 2^{n-1}
B2n\displaystyle 2^n
C2n+1\displaystyle 2^{n+1}
DNone
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Correct Answer

Option b2n\displaystyle 2^n

All Options:

  • A2n1\displaystyle 2^{n-1}
  • B2n\displaystyle 2^n
  • C2n+1\displaystyle 2^{n+1}
  • DNone

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Detailed Solution & Explanation

The question asks for the sum of the series of binomial coefficients. Although written in the question as nCr+nCr1+nCr2+\displaystyle ^nC_r + ^nC_{r-1} + ^nC_{r-2} + \dots, it represents the complete sum of binomial coefficients of order n\displaystyle n, which is:
k=0nnCk=nC0+nC1+nC2++nCn\sum_{k=0}^{n} {^nC_k} = ^nC_0 + ^nC_1 + ^nC_2 + \dots + ^nC_n
We can derive this sum using the **Binomial Theorem** expansion of (1+x)n\displaystyle (1 + x)^n:
(1+x)n=nC0x0+nC1x1+nC2x2++nCnxn(1 + x)^n = ^nC_0 x^0 + ^nC_1 x^1 + ^nC_2 x^2 + \dots + ^nC_n x^n
Substituting x=1\displaystyle x = 1 into both sides of the equation:
(1+1)n=nC0(1)0+nC1(1)1+nC2(1)2++nCn(1)n(1 + 1)^n = ^nC_0 (1)^0 + ^nC_1 (1)^1 + ^nC_2 (1)^2 + \dots + ^nC_n (1)^n
    2n=nC0+nC1+nC2++nCn\implies 2^n = ^nC_0 + ^nC_1 + ^nC_2 + \dots + ^nC_n
Therefore, the sum of all binomial coefficients of order n\displaystyle n is:
Sum=2n\text{Sum} = 2^n
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.

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