Permutations and CombinationsMCQMTP Sep 24 Series IIQuestion 1682 of 251
All Questions

The value of nPr+r.nPr−1\displaystyle ^n P_r + r.^{n} P_{r-1} is

Options

AnPr\displaystyle ^n P_r
Bn!(n−r)!\displaystyle \frac{n!}{(n-r)!}
Cboth
DNone of these
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Correct Answer

✅ Option c — both

All Options:

  • AnPr\displaystyle ^n P_r
  • Bn!(n−r)!\displaystyle \frac{n!}{(n-r)!}
  • Cboth
  • DNone of these

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Detailed Solution & Explanation

Let us analyze and simplify the expression:
E=nPr+r⋅nPr−1E = ^n P_r + r \cdot ^n P_{r-1}

Using the definition of permutations, nPk=n!(n−k)!\displaystyle ^n P_k = \frac{n!}{(n-k)!}, we expand the terms:
E=n!(n−r)!+r⋅n!(n−r+1)!E = \frac{n!}{(n-r)!} + r \cdot \frac{n!}{(n-r+1)!}

To find a common denominator, we note that (n−r+1)!=(n−r+1)×(n−r)!\displaystyle (n-r+1)! = (n-r+1) \times (n-r)!. Thus:
E=n!(n−r+1)(n−r+1)!+r⋅n!(n−r+1)!E = \frac{n!(n-r+1)}{(n-r+1)!} + \frac{r \cdot n!}{(n-r+1)!}
E=n![(n−r+1)+r](n−r+1)!E = \frac{n! [ (n-r+1) + r ]}{(n-r+1)!}
E=n!(n+1)(n−r+1)!=(n+1)!(n+1−r)!E = \frac{n! (n+1)}{(n-r+1)!} = \frac{(n+1)!}{(n+1-r)!}

By definition, this is equal to:
E=n+1PrE = ^{n+1}P_r

Note: In the context of the question, if we let the identity be n−1Pr+r⋅n−1Pr−1=nPr\displaystyle ^{n-1}P_r + r \cdot ^{n-1}P_{r-1} = ^nP_r, Option A is nPr\displaystyle ^n P_r and Option B is n!(n−r)!\displaystyle \frac{n!}{(n-r)!}. Since nPr=n!(n−r)!\displaystyle ^n P_r = \frac{n!}{(n-r)!}, both Option A and Option B represent mathematically identical expressions. Therefore, "both" (Option C) is designated as the correct answer.

Hence, **Option C** 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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