Probability of "k" successes in "n" independent trials with success rate "p" with replacement:
Binomial distribution: $nCk(p)^k(1-p)^{n-k}$
where $nCk$ is the number of different combinations that satisfies the above requirement.
$(p)^k(1-p)^{n-k}$ is the chance of getting each combination.
Probability of "k" successes in "n" independent trials with success rate "p" without replacement:
$nPk...?$
How to formulate the chance of each permutation?
Choosing 'with' and 'without' replacement enter in two different parts of the computation.
You toss a fair coin three times and want the probability of getting exactly two Heads. Then you have $X \sim \mathsf{Binom}(n = 3, p = 1/2)$ and need to compute $P(X = 2) = {3 \choose 2} (1/2)^3.$
In this simple case, equally likely outcomes are are $S = \{\text{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}\}.$ Three of these $8$ outcomes have two Heads, so the answer is $P(X = 2) = 3/8.$
According to the formula you have ${3 \choose 2} = \frac{3!}{2!\cdot 1!} = 3.$ The number of arrangements of two Heads and one Tail is found by considering three positions. You choose two positions (without replacement) for the Heads and then the Tail goes into the remaining position. So the 'without replacement' part is a combinatorial 'trick' used to discover there are two ways to get two Heads and a Tail.
The 'with replacement' part of this problem enters when we toss the coin. On each toss we obtain Head or Tail with replacement, meaning that once one Head is chosen it is possible to choose another Head on subsequent trials. So the probability of any one of the eight outcomes in my list has probability $(1/2)^3 = 1/8.$