How Many Possible NBA playoff brackets?

Important Information

The NBA playoffs is a series in basketball to determine the NBA champions. $16$ teams play in a $4$ round (round of $16$,quarter-finals,semi-finals, finals) tournament, where teams compete each round in a best-of-seven series. The first team to win $4$ games wins the series. It is therefore possible to play as few as $4$ games or as many as $7$. The bracket of the $2016$ tournament is given here to illustrate how the tournament works.

The Question

How many possible NBA playoffs brackets are there?

My Work

This process can be broken into multiple stages: choose the round-of-16 winners, the quarter finals winners, the semi-finals winners and the finals winners. Each series can involve the losing team winning $1,2$ or $3$ games.

Losing Team Wins 1

The losing team winning 1 is the number of ways to insert an L into the following sequence of W's _W_W_W_W. There are four ways to do this.

Losing Team Wins 2

The number of ways this can happen is the number of ways one can arrange 2 L's and 4 W's in such a way that an L never appears after 4 W's have appeared. This can be computed by letting $T =$ the total number of ways to arrange 4 W's and 2 L's $= \frac{6!}{4!2!}$ and letting $F =$ the number of forbidden arrangements where an L appears after 4 W's. There are two sub cases here:

One L appears after 4 W's

One L appearing after 4 W's is simply adding an L to the end of our first case so there are many ways to do this. Therefore there are 4 ways to do this.

Two L's appear after 4 W's

This simply adding two L's to the end of the 4 W's. There is exactly one way to do this WWWWLL.

Three L's appear after 4 Losses

There is exactly one way to do this WWWWLLL

Therefore, by rule of sum, there are $4+5+1 = 10$ ways for a team to win the series. This means there are 20 ways overall for either team to win a series. There are $8$ series in the first round, $4$ series in the quarter finals, $2$ series in the semis and $1$ series in the finals. By rule of product, there are $20^{8+4+2+1} = 20^{15}$ ways to win a series.

My Research

This MSE question unfortunately did not go into enough depth for me because it didn't include series results, just who won the series

My Problem

I'm looking to see if I have done this correctly, and secondly to see if there is a more simple and elegant solution than what I did.

• You might consider ${3\choose 0}$, ${4\choose 1}$, ${5\choose 2}$, ${6\choose 3}$ as the number of ways of losing a round but winning a particular number of games: they add up to ${7\choose 4}$, which is not $20$. Perhaps double this since either team could win a series. Then take your powers – Henry Aug 7 '16 at 23:31
• Don't forget, each conference plays amongst themselves - i.e., before the finals, there cannot be, say, a Spurs-Celtics series. – Sean Roberson Aug 8 '16 at 0:56
• @Henry nice solution and it also pointed out a lot of steps I missed in my solution. Thanks for the helpful comment! – Dunka Aug 8 '16 at 4:07