This is a case of combination. This is because for this case the order has no importance. Let's say students A, B, C and D were chosen for the group of 4. To choose first A, then C, D and B will gives us exactly the same group obtained of we first select student C, then B, A and D.
To solve it we can simply use the formula for combination:
[tex]\text{nCx=}\frac{n!}{x!(n-x)!}[/tex]
For this case n is 20 (the number o students) and x is 4 (the number of representatives):
[tex]\begin{gathered} 20C4=\frac{20!}{4!\cdot(20-4)!} \\ 20C4=\frac{20!}{4!\cdot16!} \end{gathered}[/tex]
20! is 20x19x18x17x16x15 and so on. We can represent it in a convenient way as: 20! = 20x19x18x17x16!:
[tex]20C4=\frac{20\cdot19\cdot18\cdot17\cdot16!}{4!\cdot16!}[/tex]
Then, the 16! in the numerator can be simplified with the 16! in the denominator:
[tex]\begin{gathered} 20C4=\frac{20\cdot19\cdot18\cdot17}{4!} \\ 20C4=\frac{20\cdot19\cdot18\cdot17}{4\cdot3\cdot2\cdot1} \\ 20C4=\frac{116280}{24}=4845 \end{gathered}[/tex]
There are 4845 different ways to choose a group of 4 among 20 people.
