>
Exams
>
Quantitative Aptitude
>
Remainders and Factorials
>
solve 7
Question:
Solve: \( 7! \) = ?
Show Hint
Factorial \( n! \) grows rapidly as \( n \) increases. \( 7! = 5040 \) is an essential value to remember for combinatorics problems.
BHU PET - 2019
BHU PET
Updated On:
Mar 25, 2025
70
120
720
5040
Hide Solution
Verified By Collegedunia
The Correct Option is
D
Solution and Explanation
Factorial of a number \( n! \) is given by: \[ n! = n \times (n-1) \times (n-2) \times ... \times 1 \] For \( 7! \): \[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \] Step-by-step calculation: \[ 7 \times 6 = 42 \] \[ 42 \times 5 = 210 \] \[ 210 \times 4 = 840 \] \[ 840 \times 3 = 2520 \] \[ 2520 \times 2 = 5040 \]
Thus, the correct answer is 5040.
Download Solution in PDF
Was this answer helpful?
0
0
Top Questions on Remainders and Factorials
What is the value of
\(0!\times5!\space ?\)
KMAT KERALA - 2023
Quantitative Aptitude
Remainders and Factorials
View Solution
What is the remainder when 496 is divided by 6?
KMAT KERALA - 2021
Quantitative Aptitude
Remainders and Factorials
View Solution
View All
Questions Asked in BHU PET exam
Which of the following is a Hermitian operator?
BHU PET - 2019
Quantum Mechanics
View Solution
Find out the next number in the series 97, 86, 73, 58, 45, (............):
BHU PET - 2019
Number Series
View Solution
In the following Venn diagram, which of the following represents the educated men but not urban?
BHU PET - 2019
Venn Diagrams
View Solution
If a particle is fixed on a rotating frame of reference, the fictitious force acting on the particle will be:
BHU PET - 2019
Rotational motion
View Solution
Given the Bessel function:
$$ J_0(x) = 1 - \frac{x^2}{2^2} + \frac{x^4}{2^2 \cdot 2^2} - \frac{x^6}{2^2 \cdot 2^2 \cdot 2^2} + \dots $$
The Bessel function $ J_1(x) $ is given by:
BHU PET - 2019
Special Functions
View Solution
View More Questions