A final examination requires that a student answer any 4 out of 6 questions. In how many ways can this be done?
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Correct Answer: Option A
Explanation:
The question will be answered in
\(^5C_4 = \frac{6!}{(6-4)!4!}\\=\frac{6!}{2!4!}\\=\frac{6\times5\times4!}{2\times1\times4!}\\=15\hspace{1mm}ways\)
The question will be answered in
\(^5C_4 = \frac{6!}{(6-4)!4!}\\=\frac{6!}{2!4!}\\=\frac{6\times5\times4!}{2\times1\times4!}\\=15\hspace{1mm}ways\)