To my friend - Aditya Kumar

All possible cases will be of the form- A 4 _ _ _ _ _ A

Case 1: Aditya gets no pass in between
A 4 (3 3 3 3 3) A - $4*3^5$ ways

Case 2: Aditya gets 1 pass in between
A 4 (1 4 3 3 3) A - The 1 signifies the case of Aditya getting the ball and the 4 to the right of it signifies the ball going to any of his 4 friends. 1 can only be at first 4 positions inside the bracket since Aditya cannot get the ball on sixth pass(fifth position inside bracket). So $4^2*3^3*4$ ways.

Case 3: Aditya gets 2 passes in between
A 4 (1 4 1 4 3) A
A 4 (1 4 3 1 4) A
A 4 (3 1 4 1 4) A - The 3 arrangements give us $4^3*3*3$ ways.

No further case is possible since we would need six positions inside the bracket to get 3 '1 4' for 3 passes to Aditya in between.

Adding up the above cases, we get $\boxed{3276}$ possible ways.

There are $4^6$ ways the ball can be passed after 6 throws, and only one choice for the seventh throw (back to Aditya). But Aditya might have the ball after 6 throws, so we need to subtract from this the number of ways Aditya can get the ball after 6 throws.

Following a similar logic, there are $4^5$ ways the ball can be passed after 5 throws, and only one choice for the seventh throw (back to Aditya). But Aditya might have the ball after 5 throws, so we need to subtract from this the number of ways Aditya can get the ball after 5 throws.

Continuing in this way, we find that the number of ways Aditya can have the ball after 7 throws is $4^6-[4^5-(4^4-[4^3-(4^2-4)])]=\boxed{3276}$ .

Fibonacci routes of type of ways of 0, 1, {1, 2, 3, 5, 8, 13, 21}:

1 2 3 4 5 6 7

1 2 3 5 8 13 18+3

4 1 4 1 4 0 0 4 1 4 1 4 0 0 0

                3   1       4   1   4   1   4   3   1   192

                    0       4   1   4   1   4   3   0   0

        3   1   4   1       4   1   4   3   1   4   1   192

            3   0   0       4   1   4   3   3   0   0   0

                3   1       4   1   4   3   3   3   1   432

                    0       4   1   4   3   3   3   0   0

3   1   4   1   4   1       4   3   1   4   1   4   1   192

            3   0   0       4   3   1   4   3   0   0   0

                3   1       4   3   1   4   3   3   1   432

                    0       4   3   1   4   3   3   0   0

    3   1   4   0   0       4   3   3   1   4   0   0   0

                3   1       4   3   3   1   4   3   1   432

                    0       4   3   3   1   4   3   0   0

        3   1   4   1       4   3   3   3   1   4   1   432

            3   0   0       4   3   3   3   3   0   0   0

                3   1       4   3   3   3   3   3   1   972

                    0       4   3   3   3   3   3   0   0

Summarized with 8 routes:

4443 192

4434 192

44333 432

4344 192

43433 432

43343 432

43334 432

433333 972

Total ways = 3276

Same question or related question which came in INMO 2014 (basketball one)

Because using the given name would give me a headache, I'm just gonna call them A, B, C, D and E, with A being the initial guy

Because B,C,D and E are similar to each others

I denote a(i) is the number of ways the ball ends up at either B, C, D or E after i turns

b(i) is the number of ways the ball return to A

Easily proven that there are 4^n ways to pass the ball in n turns

Thus we have the following:

a(n+1)=3a(n)+b(n)

b(n+1)=4a(n)

a(0)=0

b(0)=1

4a(n)+b(n)=4^n

We have the following:

a(n+1)-b(n+1) = b(n) - a(n) = (-1)^(n+1)*(a(0)-b(0)) = (-1)^n

thus a(7) = b(7) + 1

=> 4a(7)+b(7) = 5b(7)+4 = 4^7

=> b = 3276