The details as follows
Now let one of the team of North Division be X.
Case-A: So X will play a total of (4 matches *4 teams) games with other teams in the same North division = 16 matches
Case-B: X will play (3 matches * 11 teams) games with teams in the other division of the same conference = 33 matches
Case-C: X will play (2 matches * 16 teams) games with teams in the conference = 32 matches
So total matches = 81, which can be permuted in $81!$ ways.
But there is a repetition of matches like there are four matches having the same opponents when playing with the teams in the north division itself. Similarly 3 and 2 same matches in case B and C.
So we want to nullify the permutation among these same matches, and therefore we divide by $4!,\: 3! \: and \: 2!$ respectively.
And therefore total ways the TEAM-X’s matches can be scheduled = $\frac{81!}{(4!)^{4}\: (3!)^{11}\: (2!)^{16}}$