In this question, we are tossing the same coin. Consider an outcome where we get $2$ heads and $4$ tails, say $HTTTHT$ Since you are tossing the same coin, you cannot distinguish between the two heads. Similarly, you cannot distinguish between the four tails. Hence, there is only one way to get the outcome $HTTTHT$ in this order.
On the other hand, if we were to toss $6$ different colored coins to obtain $2$ heads and $4$ tails, then the outcome $\color{red}{H}\color{blue}{T}\color{green}{T}\color{orange}{T}\color{black}{H}\color{magenta}{T}$ is different from $\color{blue}{H}\color{magenta}{T}\color{red}{T}\color{black}{T}\color{orange}{H}\color{green}{T}$ even though the number of heads and tails and the order in which the heads and tails occur are the same.
We will now count the number of ways of getting the outcome $HTTTHT$ when the coins are colored differently. The first head can be obtained from any of the $6$ differently colored coins. Hence, the first position has $6$ options. Once the first position is fixed as a head, the second position i.e. the tail, can be obtained from any of the $5$ remaining coins. Fixing this, the third position i.e. again a tail, can be obtained from any of the remaining $4$ coins and so on. Hence, the total number of ways of getting $4$ heads and $2$ tails from $6$ differently colored coins in the order $HTTTHT$ is $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 6! = 720$.
Now lets get back to the original problem i.e. tossing the same coin to get $2$ heads and $4$ tails.
To look at all possible outcomes, which give us $2$ heads and $4$ tails, by the tossing of the same coin, note that all possible outcomes resulting in $2$ heads and $4$ tails by tossing the same coin $6$ times can be obtained by permuting the outcome $HTTTHT$. The total number of permutations if we do blindly is $6!$.
However, note that for instance, if the two heads were to swap its place, keeping the tails fixed and since we are tossing the same coin, we cannot distinguish between them. Hence, to take into account these are two heads from the same coin, we need to divide by $2!$.
Similarly, if all the four tails were to permute among themselves without changing the position of the two heads, we cannot distinguish this outcome from the earlier one we had. Hence, to take into account these are four tails from the same coin, we need to divide by $4!$.
Hence, the total number of ways to get $2$ heads and $4$ tails is $\dfrac{6!}{4! 2!} = \dbinom{6}{2} = \dbinom{6}{4}$