Generalised permutation and combination

Question

Suppose that a basketball league has 32 teams ,split into 2 conference of 16 teams each . Each conference is splits into 3 division .suppose that north central Division plays 4 games against each of other teams in this division , 3 games against each of the 11 remaining teams in the conference , and 2 games against each of the 16 teams in the other conference ,in how many different order can the games of one of the teams in the north central Division scheduled ?

I am unable to approach , little bit confusion in question statement ... please reply in  detail ..

Answer 1

A conference has 16 teams each
north central Division plays 4 games against each of other teams ( total 5 teams ) in this division, s0 4*(4*5/2)=4*10=40 games
3 games against each of the 11 remaining teams so 3*(5*11)=3*55=165 games
2 games against each of the 16 teams in the other conference so 2*(5*16)=2*80=160 games
so total =40+165+160=365 games will be scheduled.

Comments

thanks :)

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hey i think here somthing wrong , 

https://www.physicsforums.com/threads/permutations-matches-in-basketball-league.219838/

see this , verify your ans

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Let the 4 other teams in the NCD be x1, x2, x3, x4.

Let the 11 other teams in the conference be y1, y2, …, y11.

Let the 16 teams in the other conference be z1, z2, …, z16.

The total # of games that a team in the NCD plays is

4 . 4 + 11 . 3 + 16 . 2 = 81

The number of 81-“letter’’ sequences with:
4 each of x1, x2, x3, x4

  3 each of y1, y2, …, y11

  2 each of z1, z2, …, z16 is

81! / (4!)^4(3!)^11(2!)^16

(copied)