Answers

2014-11-07T22:50:58+08:00
This problem involves combination.
Using the formula for combination as:
_nC_r =  \frac{n!}{(n-r)!r!}
n=6
r=3
_6C_3 =  \frac{6!}{(6-3)!3!}
_6C_3 =  \frac{6(5)(4)(3)(2)(1)}{(3(2)(1))(3(2)(1))}
_6C_3 = 20
therefore there are 20 ways to which a student can select 3 books.
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