A final examination requires that a student answer any 4 out of 6 questions. In how many ways can this be done?
The question will be answered in
\(^6C_4 = \frac{6!}{(6-4)!4!}\)
=\(\frac{6!}{2!4!}\)
=\(\frac{6\times5\times4!}{2\times1\times4!}\) = 15ways.
There is an explanation video available below.
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