JAMB - Mathematics (2006 - No. 2)
A final examination requires that a student answer any 4 out of 6 questions. In how many ways can this be done?
15
20
30
45
Selitys
The question will be answered in
\(^6C_4 = \frac{6!}{(6-4)!4!}\\=\frac{6!}{2!4!}\\=\frac{6\times5\times4!}{2\times1\times4!}\\=15\hspace{1mm}ways\)
\(^6C_4 = \frac{6!}{(6-4)!4!}\\=\frac{6!}{2!4!}\\=\frac{6\times5\times4!}{2\times1\times4!}\\=15\hspace{1mm}ways\)