We can use binomial probability distribution in this case as follows:
We want number of heads as $4$ in the first $10$ tosses and number of tails as $4$ in remaining $10$ tosses.
Probability of getting $4$ heads in the first $10$ tosses can be calculated as : $\binom{10}{4}\times\left ( \frac{1}{2} \right )^{4}\times\left ( \frac{1}{2} \right )^{6}$
Probability of getting $4$ tails in the remaining $10$ tosses can be calculated as : $\binom{10}{4}\times\left ( \frac{1}{2} \right )^{4}\times\left ( \frac{1}{2} \right )^{6}$
Since the probability of getting $4$ heads the first $10$ tosses is independent of getting $4$ tails in the next $10$ tosses, we get the answer by multiplying them as :
$\rightarrow$ Probability of getting $4$ heads in the first $10$ tosses \times Probability of getting $4$ tails in the
next $10$ tosses $=$ $\frac{210\times210}{2^{20}}= \frac{44100}{1048576}=0.042$ (when rounded off to 3 decimal places)
Refer: https://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function