No. of ways to divide $2n$ seats among $3$ parties = no. of ways in which we can distribute $2n$ identical balls into $3$ distinct bins
$= {}^{2n+2}C_2.$
Now, we have to ensure that selecting any two party will get no. of seats more than the other. So, we can take the complement case where no party gets the majority of seats. Here, we have $2n$ seats and to ensure no majority one party needs $n$ seats. So, remaining $n$ seats we can distribute among $3$ parties in ${}^{n+2}C_2$ ways and the $n$ seats can go to any one among the $3$ party and thus we get $3 \times {}^{n+2}C_2$ ways.
But we over counted above. This happens when $n$ seats go to say party A and $n$ remaining seats go to party B. This case will be repeated when initial $n$ seats go to party B and remaining $n$ to party A and similarly for B and C too. So, AB-BA, AC-CA, BC-CB, 3 cases are counted extra. Excluding these our final answer will be
$ {}^{2n+2}C_2 - 3 \times {}^{n+2}C_2 + 3.$