How many 20 digit numbers have 10 even and 10 odd digits? How can I perform operations so as to get this value? Number should not have leading zeros.

A way without casework
Allowing leading $0's$, there will be $\binom{20}{10}\times 5^{20}\;$ ways
Now $\dfrac1{10}$ of cases will start with leading $0's$; subtracting these,
ans = $0.9\times\binom{20}{10}\times 5^{20}\;$ ways
