When we calculate the probability of a random selection of $3$ students being all boys, from a group of $6$ boys and $4$ girls, then we can just multiply $\mathbb P(\text{1st being boy}) \times \mathbb P(\text{2nd being boy}) \times \mathbb P(\text{3rd being boy})$ i.e. $\frac 6 {10} \times \frac 5 9 \times \frac 4 8 $.
But when we need the probability of two coin flips being TAILS in $5$ flips of a fair coin, we need to multiply the $\mathbb P(T)\times\mathbb P(T)\times\mathbb P(H)\times\mathbb P(H)\times\mathbb P(H)\times{5\choose 2}$.
Why do we need to multiply the probabilities by the number of combinations in the coin flips problem but not in the first?
Because there is only one way to get three boys. If you were calculating the probability of getting two boys and one girl, you could do $\frac 6{10} \cdot \frac 59 \frac 48$, which chooses two boys first, then one girl, then multiply by $3$ because there are three orders you can choose the three children.