Respuesta :
We have to find the value of 9P3 which means the Permutations of 9 objects taken 3 at a time.
The formula of permutations is:
[tex]nPr = \frac{n!}{(n-r)!} [/tex]
In the given case n=9 and r=3. Using these values, we get:
[tex]9P3= \frac{9!}{(9-3)!} \\ \\ = \frac{9!}{6!} \\ \\ = \frac{9*8*7*6!}{6!} \\ \\ =9*8*7 \\ \\ =504 [/tex]
Thus 9P3 is equal to 504.
9! is expanded above using the following factorial formula:
n! = n(n-1)(n-2)(n-3)....3.2.1
The formula of permutations is:
[tex]nPr = \frac{n!}{(n-r)!} [/tex]
In the given case n=9 and r=3. Using these values, we get:
[tex]9P3= \frac{9!}{(9-3)!} \\ \\ = \frac{9!}{6!} \\ \\ = \frac{9*8*7*6!}{6!} \\ \\ =9*8*7 \\ \\ =504 [/tex]
Thus 9P3 is equal to 504.
9! is expanded above using the following factorial formula:
n! = n(n-1)(n-2)(n-3)....3.2.1
Answer:
The value of the expression is 504.
Step-by-step explanation:
We are given the expression [tex]9P3[/tex].
Now, [tex]9P3[/tex] represents the permutations of 9 objects which are taken 3 at a time.
Also, its value is given by,
[tex]9P3=\dfrac{9!}{(9-3)!}\\\\9P3=\dfrac{9!}{6!}[/tex],
where ! represents the factorial given by,
[tex]n!=n(n-1)(n-2)...3.2.1[/tex]
Thus, evaluating the expression, we get,
[tex]9P3=\dfrac{9!}{6!}\\\\9P3=\dfrac{9\times 8\times 7\times 6!}{6!}\\\\9P3=9\times 8\times 7\\\\9P3=504[/tex]
