# Question 0dedf

Dec 12, 2016

$70$

#### Explanation:

Assuming the $8$ objects are distinct, then this is equivalent to the number of ways of choosing $4$ objects from the group of $8$, as if that is the first group, then the remaining $4$ is decided as the second group.

The number of ways of choosing $k$ objects from a set of $n$ objects can be calculated as

((n),(k)) = (n!)/(k!(n-k)!)

(read as $n$ choose $k$)

Applying this to the given question, we have the number of ways of choosing a set of $4$ objects from a set of $8$ as

((8),(4)) = (8!)/(4!(8-4)!)

=(8*7*6*...*3*2*1)/((4*3*2*1)(4*3*2*1)#

$= \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}$

$= 70$