A school district has 2 teaching positions to fill and there are 8 applicants to choose from. How many different possibilities are there?

I did the previous one for you.

Do be successful in this topic you simply MUST know how to do this type of trivial question.
They don't come any easier than this one.

Show me the steps you would take to solve this problem.

one way i got 8!/2!*(8-2)!

8!/(2!*6!)
8!=40320
2!=2
6!=720
40320/(2*720 )=40320/1440=28 or another 7*8=56

can somebody check this problem

thank you.

To find the number of different possibilities, we can use the concept of combinations.

In this scenario, we need to choose 2 applicants out of 8. We can calculate this using the formula for combinations, which is:
C(n, r) = n! / (r!(n-r)!)

Where:
C(n, r) represents "n choose r" (the number of combinations of n items taken r at a time),
n! represents the factorial of n (the product of all positive integers from 1 to n),
r! represents the factorial of r, and
(n-r)! represents the factorial of (n-r).

In this case, we have 8 applicants (n = 8) and we need to choose 2 (r = 2). Plugging these values into the formula, we get:
C(8, 2) = 8! / (2!(8-2)!)

Now, let's calculate this step by step:
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320
2! = 2 × 1 = 2
(8-2)! = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Plugging these values back into the formula, we have:
C(8, 2) = 40,320 / (2 × 720)

Simplifying the expression, we get:
C(8, 2) = 40,320 / 1,440

Finally, dividing these numbers, we find the total number of different possibilities:
C(8, 2) = 28

Therefore, there are 28 different possibilities for filling the 2 teaching positions from the 8 applicants.