Permutation or combination?

Question

(a) Find the total number of different selections of 4 apples from 12 apples.

The answer is 495. So I thought the solution would be something like this:

12C4 = 495.

I don't why it is 12C4.. Why can't it be 12P4...

In here, the arrangement is important?

Ahhhh.. I'm confused.

(b) If 3 of these 12 apples are bruised, and a random selection of 4 apples is made, find the probability that the selection will contain precisely one bruised apple.

For this, I thought of doing something like this:

(3/4 * 3/4 * 1/4 * 1/4) * 4

The 3/4 I got from 9/12 (which represent the apples that are NOT bruised.)

The 1/4 I got from 3/12 (which represent the bruised apples.)

I multiply by 4 because the arrangements of these apples can be changed.

The correct answer for part b is 28/55.

Please help me... I don't know what I'm doing...

Answer 1

a) it's 12C4 because order does not matter

b) first bruised apple = 3/12

second good apple = 9/11 --- this is because after you get out one bruised apple, there are 11 apples left and 9 of them are good apples

third good apple = 8/10

fourth good apple = 7/9

4 (3/12) (9/11) (8/10) (7/9) = 28/55

you can use other method

P = (3C1)(9C3) / (12C4) = 28/55