If nP = 720 and nC = 120 , then n is

Option a: 3. Solution: We are given the values for permutations and combinations for the same parameters n and r : nPr = 720 and nCr = 120 We know the fundamental relationship between permutations and combinations: nPr = r! × nCr Substituting the given values: 720 = r! × 120 r! = 720/120 = 6 Since 3! = 3 × 2 × 1 = 6 , we find: r = 3 Now, using the permutation formula for nP3 = 720 : nP3 = n!/(n-3)! = n(n-1)(n-2) = 720 We factor 720 into three consecutive positive integers: 720 = 10 × 9 × 8 Comparing this with n(n-1)(n-2) = 720 , we get: n = 10 Since the question choices are 3, 4, 6, 5, the question is asking for the value of r which is 3. Hence, Option A is the correct answer.

If $^nP = 720$ and $^nC = 120$, then $n$ is

The correct answer is Option a: 3. We are given the values for permutations and combinations for the same parameters $n$ and $r$: $$^nP_r = 720 \quad \text{and} \quad ^nC_r = 120$$ We know the fundamental relationship between permutations and combinations: $$^nP_r = r! \times ^nC_r$$ Substituting the given values: $$720 = r! \times 120$$ $$r! = \frac{720}{120} = 6$$ Since $3! = 3 \times 2 \times 1 = 6$, we find: $$r = 3$$ Now, using the permutation formula for $^nP_3 = 720$: $$^nP_3 = \frac{n!}{(n-3)!} = n(n-1)(n-2) = 720$$ We factor 720 into three consecutive positive integers: $$720 = 10 \times 9 \times 8$$ Comparing this with $n(n-1)(n-2) = 720$, we get: $$n = 10$$ Since the question choices are 3, 4, 6, 5, the question is asking for the value of $r$ which is 3. Hence, **Option A** is the correct answer.

Permutations and CombinationsMCQPYQ Nov 18Question 1684 of 251

All Questions

If $\displaystyle ^nP = 720$ and $\displaystyle ^nC = 120$ , then $\displaystyle n$ is

Options

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 3

All Options:

Detailed Solution & Explanation

We are given the values for permutations and combinations for the same parameters

\displaystyle n

and

\displaystyle r

^nP_r = 720 \quad \text{and} \quad ^nC_r = 120

We know the fundamental relationship between permutations and combinations:

^nP_r = r! \times ^nC_r

Substituting the given values:

720 = r! \times 120

r! = \frac{720}{120} = 6

Since

\displaystyle 3! = 3 \times 2 \times 1 = 6

, we find:

r = 3

Now, using the permutation formula for

\displaystyle ^nP_3 = 720

^nP_3 = \frac{n!}{(n-3)!} = n(n-1)(n-2) = 720

We factor 720 into three consecutive positive integers:

720 = 10 \times 9 \times 8

Comparing this with

\displaystyle n(n-1)(n-2) = 720

, we get:

n = 10

Since the question choices are 3, 4, 6, 5, the question is asking for the value of

\displaystyle r

which is 3.

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

Ready to Master Permutations and Combinations?

Practice all 251 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free

If nP=720\displaystyle ^nP = 720nP=720 and nC=120\displaystyle ^nC = 120nC=120, then n\displaystyle nn is