Basketball and Combinatorics

Let $x_{n}$ be the number of ways in which A can get back the ball after n passes. Let $y_{n}$ be the number of ways in which the ball goes back to a fixed person other than A after $n$ passes. We need to find $x_7$ . Then:

$x_{n}$ = $3y_{n-1}$

Also:

$y_{n}$ = $x_{n-1}$ + $2y_{n-1}$

We also know that $x_1 = 0$ , $x_2 = 3$ , $y_1 = 1$ , and $y_2 = 2$ .

Then just eliminate $y_n$ and $y_{n-1}$ to get the following recurrence relation:

$x_n$ = $3x_{n-1} + 2x_n$ .

Using this relation, we get:

$x_3 = 6$ ;

$x_4 = 21$ ;

$x_5 = 60$ ;

$x_6 = 183$ ;

$x_7 = \boxed{546}$

Let A, B, C, D be the four players. We establish a bijection that the answer is the same as number of ways of writing the sequence of 8 letters starting and ending with A, with no consecutive places occupied by the same letter.
Like, $A, B, C, A, B, D, B, A$ is a permissible sequence but $A, B, B, A, B, D, B, A$ and $A, B, C, A, B, D, B, C$ are not.

CASE 1: No A's in between like $A, B, C, D, B, D, B, A$ .
There are $2^{5}×3$ such arrangements.

CASE 2:1 A in between like $A, B, C, A, B, D, B, A$ .
There are $4×3^{2}×2^{3}$ such arrangements.

CASE 3: 2 A's in between like $A, B, C, A, B, A, B, A$ .
There are $3×3^{3}×2$ such arrangements.

We don't have a case four since we can't have 3 A's in between, otherwise two of them will be adjacent.

Now adding all the cases we get the answer "546".