Question #0dedf

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Question #0dedf

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Answer 1 · 2016-12-12T03:56:33

$70$

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

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

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

$=(8*7*6*5)/(4*3*2*1)$

$=70$

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Question #0dedf

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