Permutations and CombinationsMCQPYQ Jan 21Question 1694 of 249
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nCp+2⋅nCp−1+nCp−2=?\displaystyle ^nC_p + 2 \cdot ^nC_{p-1} + ^nC_{p-2} =?

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option b — n+2Cp\displaystyle ^{n+2}C_p

All Options:

  • An+2Cp\displaystyle ^{n+2}C_p
  • Bn+2Cp\displaystyle ^{n+2}C_p
  • Cn+1Cp\displaystyle ^{n+1}C_p
  • Dn+1Cp−1\displaystyle ^{n+1}C_{p-1}

Detailed Solution & Explanation

We want to simplify the expression:
E=nCp+2⋅nCp−1+nCp−2E = ^nC_p + 2 ⋅\displaystyle \cdot ^nC_{p-1} + ^nC_{p-2}
We can split the middle term 2⋅nCp−1\displaystyle 2 ⋅\displaystyle \cdot ^nC_{p-1} into nCp−1+nCp−1\displaystyle ^nC_{p-1} + ^nC_{p-1}:
E=(nCp+nCp−1)+(nCp−1+nCp−2)E = \left(^nC_p + ^nC_{p-1}\right) + \left(^nC_{p-1} + ^nC_{p-2}\right)
Recall Pascal's Identity: nCr+nCr−1=n+1Cr\displaystyle ^nC_r + ^nC_{r-1} = ^{n+1}C_r.
Applying Pascal's Identity to both terms in parentheses:
1. nCp+nCp−1=n+1Cp\displaystyle ^nC_p + ^nC_{p-1} = ^{n+1}C_p
2. nCp−1+nCp−2=n+1Cp−1\displaystyle ^nC_{p-1} + ^nC_{p-2} = ^{n+1}C_{p-1}
Substituting these back into the expression:
E=n+1Cp+n+1Cp−1E = ^{n+1}C_p + ^{n+1}C_{p-1}
Applying Pascal's Identity once more on the simplified expression:
E=(n+1)+1Cp=n+2CpE = ^{(n+1)+1}C_p = ^{n+2}C_p
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

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