permutation

how many way we can arange thing if we care about the order? $$_kP_n = \frac{n!}{(n-k)!}$$ example: how many way 5 person can be seated to seat numbered from 1-3 = $_3P_5$ answer = 5 way for the first seat, 4 way for the 2nd seat, 3 way for the 3rd seat $$\frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 * 4 * 3 = 60$$

combination

how many way we can get item/thing if we care about the order? $$_kC_n = \frac{n!}{k!(n-k)!}$$ example: if we have 5 ball numbered 1-5, how many possibilities we can get if we take the ball blindly and randomly? answer = 3 way to take the ball (5 * 4 * 3) and since we don’t care about the order we divide it by the number of possible order of each, since we double counting the balls, so we divide it by (3 * 2 * 1) $$_3C_5 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 * 4 * 3}{3 * 2 *1} = 10$$

References

[1] https://www.khanacademy.org/math/statistics-probability/counting-permutations-and-combinations/