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Suppose that a basketball league has $32$ teams, split into two conferences of $16$ teams each. Each conference is split into three divisions. Suppose that the North Central Division has five teams. Each of the teams in the North Central Division plays four games against each of the other teams in this division, three games against each of the $11$ remaining teams in the conference, and two games against each of the $16$ teams in the other conference. In how many different orders can the games of one of the teams in the North Central Division be scheduled?

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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 = $\frac{81!}{(4!)^{4}\: (3!)^{11}\: (2!)^{16}}$

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admin asked May 1, 2020
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