1
$\begingroup$

I have a the link below to show the Bracket I was given(the teams are different, that should not make a difference) There is 16 teams on the Sheet i was given, with what it looks like if the brackets are correct, 13 games. I need to find 2 answers

1). How many possible wasy are there for how the entire NBA playoffs could play out?

2). How many possible ways are there for how the entire NBA plays could play out WITH the ATLANTA HAWKS winning?

I believe this is a permutation but i do not know how to find the answers please help.

http://www.scoresreport.com/wp-content/uploads/2010/04/nba-playoffs-bracket.jpg

  • 1
    I'm pretty sure the (practical) answer to the second question is ZERO.2012-10-15

1 Answers 1

2

There should be $15$ matches, not $13.$ Hint: Each match has $2$ possible results. If the Hawks win, how many of the matches are determined. How many are left?

  • 0
    so is it the permutation of 16!/1!? and im not sure as far as the matches left..2012-10-14
  • 1
    @John: It’s not permutations at all: it’s something much simpler. You have a sequence of $15$ ‘choices’, and each ‘choice’ can be made in $2$ ways; how many different ways are there to make the sequence of ‘choices’? For your other question, if the Hawks are champions, they must win every one of their matches. How many matches is that? How many other matches are there that can go either way?2012-10-14
  • 0
    Okay, the way the teacher explained it, it sounded like permutations. I wasn't sure. So, it would be 15x2=30, and from the bracket i see it must be 3 brackets for the hawks to win so 6games? and then 12 are left so 12x2=24?2012-10-14
  • 1
    @John: Not $15\cdot2=30$. You need the multiplication rule, sometimes called the Chinese menu rule: you **multiply** the numbers of choices to get the total number of ways that a sequence of choices can be made. In this case that gives you $2^{15}$ different ways for the tournament to play out. The Hawks must beat **four** opponents, not three.2012-10-14
  • 0
    So 32768 for the first answer. Now if the hawks defeat four opponents, would that mean I subtract 15-4 to get 11 then take 2^11 using the multiplication rule(Chinese menu rule) to get the product of 2048? Where would I include the 4 games played by the Hawks? would I have to due 2^4 and add it in. Sorry, I don't really know how to do this, and thank you for helping.2012-10-14
  • 0
    @John: That is right. The four games that the Hawks win now have only one choice, so it would be $2^{11}\cdot 1^4$, but the $1^4$ doesn't change anything. You accounted for those four games with your $15-4=11$2012-10-15