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:
1 or 2
Step-by-step explanation:
Write down the numbers that go into each one
12: 1, 2, 3, 4, 6, 12
14: 1, 2, 7, 14
16: 1, 2, 3, 6, 8, 16
The numbers that all three have in common are 1 and 2
B and c boiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Answer:
when x = 0, the answer is 0
when x = 3, the answer is 24
when x = 4, the answer is 32
Answer:
x=2
Step-by-step explanation:
let the number be x
9x-8=6+2x
9x-2x=6+8
7x=14
x=14/7
x=2
