Answer:
2
Step-by-step explanation:
If you subtract 8from20 there's 11 memeber student councilors left which then you subtract another 8from the remaining 11 students and there would be 3 student councilors left which means
there can only be 2.5 different committees because you can't subtract 3 from 8
Answer:
125790 different committees can be chosen.
Step-by-step explanation:
The principal would like to assemble a committee of 8 students from the 20-member student council.
We need to find How many different committees can be chosen?
Since the order is not important, so we use Combinations to solve the question.
The formula used is: [tex]^nC_r=\frac{n!}{(n-r)!r!} \\[/tex]
In the question given: we have n = 20 and r = 8
Putting values and finding the number of ways committees can be chosen:
[tex]^nC_r=\frac{n!}{(n-r)!r!} \\\\^{20}C_8=\frac{20!}{(20-8)!8!} \\^{20}C_8=\frac{20*19*18*17*16*15*14*13*12!}{12!8!} \\^{20}C_8=\frac{20*19*18*17*16*15*14*13}{8!} \\^{20}C_8=\frac{20*19*18*17*16*15*14*13}{8*7*6*5*4*3*2*1} \\^{20}C_8=\frac{5079110400}{40320}\\^{20}C_8=125970[/tex]
So, 125790 different committees can be chosen.
