three out of nine students in the computer club are getting prizes for first, second, and third place in a…

three out of nine students in the computer club are getting prizes for first, second, and third place in a competition. how many ways can first, second, and third place be assigned? $_9p_3=\frac{9!}{(9 - 3)!}$\no 3\no 84\no 504\no 2048

three out of nine students in the computer club are getting prizes for first, second, and third place in a competition. how many ways can first, second, and third place be assigned? $_9p_3=\frac{9!}{(9 - 3)!}$\no 3\no 84\no 504\no 2048

Answer

Explanation:

Step1: Recall permutation formula

The permutation formula is ${n}P{r}=\frac{n!}{(n - r)!}$, where $n$ is the total number of items and $r$ is the number of items to be selected and arranged. Here $n = 9$ and $r=3$, so ${9}P{3}=\frac{9!}{(9 - 3)!}$.

Step2: Expand factorials

We know that $n!=n\times(n - 1)\times\cdots\times1$. So $9! = 9\times8\times7\times6\times5\times4\times3\times2\times1$ and $(9 - 3)!=6!=6\times5\times4\times3\times2\times1$. Then ${9}P{3}=\frac{9!}{6!}=\frac{9\times8\times7\times6!}{6!}$.

Step3: Simplify the expression

Cancel out the $6!$ terms in the numerator and denominator. We get ${9}P{3}=9\times8\times7=504$.

Answer:

504