If nP6 = 336 and nCr = 56 , then n and r will be

Option b: (8,3). Solution: Let us first analyze the equations printed in the question: nP6 = 336 nCr = 56 If n = 7 , then 7P6 = 7! = 5040 ≠ 336 . This indicates a typographical error where nPr was printed as nP6 . Let us solve the correct intended equations: nPr = 336 nCr = 56 Using the fundamental relationship nPr = r! × nCr , we substitute the values: 336 = r! × 56 r! = 336/56 = 6 Since 3! = 6 , we find: r = 3 Now we substitute r = 3 into the combination equation: nC3 = 56 (n-1)(n-2)/3! = 56 n(n-1)(n-2) = 56 × 6 = 336 We need three consecutive positive integers whose product is 336. Testing values: 8 × 7 × 6 = 336 Thus, we find n = 8 . Therefore, the correct values are n = 8 and r = 3 , which corresponds to Option B. The textbook key incorrectly lists Option C ( (7,4) ) as correct. We have shown the correct derivation of (8,3) . Hence, Option B is the correct answer.

If $^nP_6 = 336$ and $^nC_r = 56$, then n and r will be

The correct answer is Option b: $(8,3)$. Let us first analyze the equations printed in the question: $$^nP_6 = 336 \\quad \\text{and} \\quad ^nC_r = 56$$ If $n = 7$, then $^7P_6 = 7! = 5040 \\neq 336$. This indicates a typographical error where $^nP_r$ was printed as $^nP_6$. Let us solve the correct intended equations: $$^nP_r = 336 \\quad \\text{and} \\quad ^nC_r = 56$$ Using the fundamental relationship $^nP_r = r! \\times ^nC_r$, we substitute the values: $$336 = r! \\times 56$$ $$r! = \\frac{336}{56} = 6$$ Since $3! = 6$, we find: $$r = 3$$ Now we substitute $r = 3$ into the combination equation: $$^nC_3 = 56$$ $$\\frac{n(n-1)(n-2)}{3!} = 56$$ $$n(n-1)(n-2) = 56 \\times 6 = 336$$ We need three consecutive positive integers whose product is 336. Testing values: $$8 \\times 7 \\times 6 = 336$$ Thus, we find $n = 8$. Therefore, the correct values are $n = 8$ and $r = 3$, which corresponds to Option B. The textbook key incorrectly lists Option C ($(7,4)$) as correct. We have shown the correct derivation of $(8,3)$. Hence, **Option B** is the correct answer.

Permutations and CombinationsMCQMTP May 19Question 1637 of 251

All Questions

If $\displaystyle ^nP_6 = 336$ and $\displaystyle ^nC_r = 56$ , then n and r will be

Options

\displaystyle (3,2)

\displaystyle (8,3)

\displaystyle (7,4)

Dnone of these

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

Correct Answer

✅ Option b — $\displaystyle (8,3)$

All Options:

A $\displaystyle (3,2)$
B $\displaystyle (8,3)$
C $\displaystyle (7,4)$
Dnone of these

Detailed Solution & Explanation

Let us first analyze the equations printed in the question:

^nP_6 = 336 \\quad \\text{and} \\quad ^nC_r = 56

\displaystyle n = 7

, then

\displaystyle ^7P_6 = 7! = 5040 \\neq 336

. This indicates a typographical error where

\displaystyle ^nP_r

was printed as

\displaystyle ^nP_6

.

Let us solve the correct intended equations:

^nP_r = 336 \\quad \\text{and} \\quad ^nC_r = 56

Using the fundamental relationship

\displaystyle ^nP_r = r! \\times ^nC_r

, we substitute the values:

336 = r! \\times 56

r! = \\frac{336}{56} = 6

Since

\displaystyle 3! = 6

, we find:

r = 3

Now we substitute

\displaystyle r = 3

into the combination equation:

^nC_3 = 56

\\frac{n(n-1)(n-2)}{3!} = 56

n(n-1)(n-2) = 56 \\times 6 = 336

We need three consecutive positive integers whose product is 336. Testing values:

8 \\times 7 \\times 6 = 336

Thus, we find

\displaystyle n = 8

.
Therefore, the correct values are

\displaystyle n = 8

and

\displaystyle r = 3

, which corresponds to Option B. The textbook key incorrectly lists Option C (

\displaystyle (7,4)

) as correct. We have shown the correct derivation of

\displaystyle (8,3)

.
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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If nP6=336\displaystyle ^nP_6 = 336nP6​=336 and nCr=56\displaystyle ^nC_r = 56nCr​=56, then n and r will be