This is a combination problem.
Given:
12 students
3 groups consisting of 4 students.
Mark can't be in the first group.
The combination formula that I used is: n! / r!(n-r)!
where: n = number of choices ; r = number of people to be chosen.
This is the formula I used because the order is not important and repetition is not allowed.
Since Mark can't be considered in the first group, the value of n would be 11 instead of 12. value of r is 4.
numerator: n! = 11! = 39,916,800
denominator: r!(n-r)! = 4!(11-4)! = 4!*7! = 120,960
Combination = 39,916,800 / 120,960 = 330
There are 330 ways that the instructor can choose 4 students for the first group
Answer:
4
Step-by-step explanation:
log(50×50)×log(2×2)
log (2500)×log(4)
log(2500×4)
log(10000)
log(10^4)
=4
Answer:
d
Step-by-step explanation:
Answer:
(7, 3)
Step-by-step explanation:
get rid of the y by adding:
x=-2
+x=16
then add the equalities:
2x=14
divide by 2:
x=7
then subsitute the x in to find y!
7+3y=16
3y=16-7
3y=9
y=3
