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 : 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.

Permutations and CombinationsMCQPYQ Nov 18Question 1684 of 249

All Questions

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

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

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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

We know the fundamental relationship between permutations and combinations:

^nP_r = r!

Substituting the given values:

720 = r!

r! =

Since $3! = 3

\displaystyle \times

\displaystyle \times

1 = 6

\displaystyle , we find:<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>r</mi><mo>=</mo><mn>3</mn></mrow><annotation encoding="application/x-tex">r = 3</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">r</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">3</span></span></span></span></span><br><br>Now, using the permutation formula for

^nP_3 = 720

\displaystyle :<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mrow></mrow><mi>n</mi></msup><msub><mi>P</mi><mn>3</mn></msub><mo>=</mo><mfrac><mrow><mi>n</mi><mo stretchy="false">!</mo></mrow><mrow><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>3</mn><mo stretchy="false">)</mo><mo stretchy="false">!</mo></mrow></mfrac><mo>=</mo><mi>n</mi><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo><mo>=</mo><mn>720</mn></mrow><annotation encoding="application/x-tex">^nP_3 =

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

\displaystyle , we get:<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>n</mi><mo>=</mo><mn>10</mn></mrow><annotation encoding="application/x-tex">n = 10</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">n</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">10</span></span></span></span></span><br><br>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.

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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Permutations vs Combinations: The Quantitative Aptitude Guide

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