
How do you evaluate ${}^{9}{{p}_{4}}$?
Answer
555k+ 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.
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