A fair coin is tossed 10 times. The probability of getting as many heads in the first 5 tosses as in the last 5 tosses is

$\mathrm{n}\left(\mathrm{S}\right)=2^{10}$

E=getting as many heads in the 5 tosses as in the last 5 tosses 

$\mathrm{n}\left(\mathrm{E}\right)={}^5{\mathrm{C}}_0 {}^5{\mathrm{C}}_0 +{}^5{\mathrm{C}}_1 {}^5{\mathrm{C}}_1 +....+{}^5{\mathrm{C}}_5 {}^5{\mathrm{C}}_5 ={}^{10}{\mathrm{C}}_5$

$\Rightarrow \mathrm{P}\left(\mathrm{E}\right)=\frac{{}^{10}{\mathrm{C}}_5}{2^{10}}$