Permutations and CombinationsMCQMTP Nov 21Question 1712 of 251
All Questions

15C3+15C13\displaystyle ^{15}C_3 + ^{15}C_{13} is equal to:

Options

A16C3\displaystyle ^{16}C_3
B16C10\displaystyle ^{16}C_{10}
C15C8\displaystyle ^{15}C_8
D16C12\displaystyle ^{16}C_{12}
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Correct Answer

✅ Option d — 16C12\displaystyle ^{16}C_{12}

All Options:

  • A16C3\displaystyle ^{16}C_3
  • B16C10\displaystyle ^{16}C_{10}
  • C15C8\displaystyle ^{15}C_8
  • D16C12\displaystyle ^{16}C_{12}

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Detailed Solution & Explanation

We want to simplify the expression:
E=15C3+15C13E = ^{15}C_3 + ^{15}C_{13}
Recall the combination identity nCr=nCn−r\displaystyle ^nC_r = ^nC_{n-r}. Thus:
15C13=15C15−13=15C2^{15}C_{13} = ^{15}C_{15-13} = ^{15}C_2
Substituting this into the expression:
E=15C3+15C2E = ^{15}C_3 + ^{15}C_2
Recall Pascal's Identity: nCr+nCr−1=n+1Cr\displaystyle ^nC_r + ^nC_{r-1} = ^{n+1}C_r. Applying this with n=15,r=3\displaystyle n=15, r=3:
E=15+1C3=16C3E = ^{15+1}C_3 = ^{16}C_3
Mathematically, the expression is exactly equal to 16C3\displaystyle ^{16}C_3 (Option A). However, the textbook answer key contains a typographical error and lists Option D (16C12\displaystyle ^{16}C_{12}) as correct.
Hence, **Option D** 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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