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

8 0

4 years ago

Read 2 more answers

Solve the system equation

den301095 [7]

<h2>Answer:</h2>

We will use substitution to solve this system for x.

$x - 2y = 27\\y = 5x\\\\x - 2(5x) = 27\\\\x - 10x = 27\\\\-9x = 27\\\\x = -3$

Now that we know what x is, we can determine y by pluging in -3 for x in one of the original equations. I will use the first one.

$-3 - 2y = 27\\\\-2y = 30\\\\y = -15\\$

We now have our solution.

D. $(-3, -15)$

We can check our answer by pluging the solution into one of the original equations to see if both sides equal each other. I will use the first equation.

$-3 - 2(-15) = 27\\\\-3 + 30 = 27\\\\27 = 27$

Since the equation checks out, our solution is correct.

4 0

3 years ago