Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.5k views
in Mathematics by (115k points)
closed by
If 2nP3 = 100 nP2, then the value of n is:
1. 1
2. 50
3. 13
4. 4

1 Answer

0 votes
by (114k points)
selected by
 
Best answer
Correct Answer - Option 3 : 13

Concept:

Permutations and Combinations:

  • Pr = \(\rm \frac{n!}{(n - r)!}\).
  • n! = 1 × 2 × 3 × ... × n.
  • 0! = 1.

 

Calculation:

Expanding the expression on both sides of the equation, we get:

2nP3 = 100 nP2

⇒ \(\rm \frac{2n!}{(2n - 3)!}=100\frac{n!}{(n - 2)!}\)

⇒ (2n)(2n - 1)(2n - 2) = 100(n)(n - 1)

⇒ 2n - 1 = 25, n ≠ 1.

⇒ n = 13.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...