Respuesta :
If the order doesn't matter the it's
[tex]C(20,4)=\dfrac{20!}{4!16!}=\dfrac{17\cdot18\cdot19\cdot20}{2\cdot3\cdot4}=4845[/tex]
[tex]C(20,4)=\dfrac{20!}{4!16!}=\dfrac{17\cdot18\cdot19\cdot20}{2\cdot3\cdot4}=4845[/tex]
Answer:
Different ways can these students be selected = 4845
Step-by-step explanation:
Number of combinations possible from n number of population if r people are selected is [tex]^nC_r[/tex]
Here n = 20 and r = 4
Different ways can these students be selected = [tex]^{20}C_4[/tex]
[tex]^{20}C_4=\frac{20\times 19\times 18\times 17}{1\times 2\times 3\times 4}=4845[/tex]
Different ways can these students be selected = 4845