
How do you evaluate ${}^{9}{{p}_{4}}$?
Answer
549k+ views
Hint:To solve this problem, we will use the formula of permutation. ${}^{n}{{p}_{r}}=\dfrac{n!}{(n-r)!}$ is the formula of permutation which will help us to solve this problem. n number of permutations within r time is determined by ${}^{n}{{p}_{r}}=\dfrac{n!}{(n-r)!}$ . So let’s see how we can solve this problem.
Complete Step by Step Solution:
In the given question we have to evaluate ${}^{9}{{p}_{4}}$ . So here we will use the formula of permutation that is ${}^{n}{{p}_{r}}=\dfrac{n!}{(n-r)!}$ where $n! = n (n-1) (n-2)…… \times 3 \times 2 \times 1$.
So here the value of n is 9 and the value of r is 4.
$\Rightarrow {}^{9}{{p}_{4}}=\dfrac{9!}{(9-4)!}$
$\Rightarrow {}^{9}{{p}_{4}}=\dfrac{9!}{5!}$
So here we will solve the factorial
$\Rightarrow \dfrac{9!}{5!}=\dfrac{9\times 8\times 7\times 6\times 5!}{5!}$
$\Rightarrow \dfrac{9!}{5!}=\dfrac{9\times 8\times 7\times 6\times \not{5}\not{!}}{\not{5}\not{!}}=9\times 8\times 7\times 6$
Now, we will calculate the numerator
= 3024
$\therefore {}^{9}{{p}_{4}}=3024$
Additional Information:
The act of arranging the objects or numbers in some specific sequence or order is known as permutation. And the combination is the way of selecting objects such that the selection order does not affect the combination.
Note:
In the given problem the factorial “!” is the important part. For solving the factorial, we have to subtract and multiply the number until we reach 1. Like if we have a number 10 and we have to find the factorial of 10. Then $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800$ will be the factorial of 10. So, as you have seen that we will stop once we reach 1.
Complete Step by Step Solution:
In the given question we have to evaluate ${}^{9}{{p}_{4}}$ . So here we will use the formula of permutation that is ${}^{n}{{p}_{r}}=\dfrac{n!}{(n-r)!}$ where $n! = n (n-1) (n-2)…… \times 3 \times 2 \times 1$.
So here the value of n is 9 and the value of r is 4.
$\Rightarrow {}^{9}{{p}_{4}}=\dfrac{9!}{(9-4)!}$
$\Rightarrow {}^{9}{{p}_{4}}=\dfrac{9!}{5!}$
So here we will solve the factorial
$\Rightarrow \dfrac{9!}{5!}=\dfrac{9\times 8\times 7\times 6\times 5!}{5!}$
$\Rightarrow \dfrac{9!}{5!}=\dfrac{9\times 8\times 7\times 6\times \not{5}\not{!}}{\not{5}\not{!}}=9\times 8\times 7\times 6$
Now, we will calculate the numerator
= 3024
$\therefore {}^{9}{{p}_{4}}=3024$
Additional Information:
The act of arranging the objects or numbers in some specific sequence or order is known as permutation. And the combination is the way of selecting objects such that the selection order does not affect the combination.
Note:
In the given problem the factorial “!” is the important part. For solving the factorial, we have to subtract and multiply the number until we reach 1. Like if we have a number 10 and we have to find the factorial of 10. Then $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800$ will be the factorial of 10. So, as you have seen that we will stop once we reach 1.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Difference Between Prokaryotic Cells and Eukaryotic Cells

