Question

evaluate $_9c_4$ and $_8p_3$.

$_9c_4=

$_8p_3=$

Explanation:

Step1: Recall combination formula

The combination formula is $_{n}C_{r}=\frac{n!}{r!(n - r)!}$. For $_{9}C_{4}$, $n = 9$ and $r=4$.
$_{9}C_{4}=\frac{9!}{4!(9 - 4)!}=\frac{9!}{4!5!}$

Step2: Expand factorials

$9! = 9\times8\times7\times6\times5!$, so $_{9}C_{4}=\frac{9\times8\times7\times6\times5!}{4!5!}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126$

Step3: Recall permutation formula

The permutation formula is $_{n}P_{r}=\frac{n!}{(n - r)!}$. For $_{8}P_{3}$, $n = 8$ and $r = 3$.
$_{8}P_{3}=\frac{8!}{(8 - 3)!}=\frac{8!}{5!}$

Step4: Expand factorials

$8! = 8\times7\times6\times5!$, so $_{8}P_{3}=\frac{8\times7\times6\times5!}{5!}=8\times7\times6 = 336$

Answer:

$_{9}C_{4}=126$
$_{8}P_{3}=336$