The Ultimate Football Challenge

Since Messi receives the ball at $10^{th}$ pass, therefore the ball is possessed by $11$ players during the course of the goal.

Let the sequence of players to possess the ball be: $\{P_1, P_2, P_3, P_4, P_5, P_6, P_7, P_8, P_9, P_{10}, P_{11}\}$ .

In order to find the total ways to pass the ball, effectively we need to find permutations of $8$ players $\{P_3, P_4, P_5, P_6, P_7, P_8, P_9, P_{10}\}$ given that $P_2$ can have $10$ choices and $P_{10} \neq$ Messi .

The sequence $\{P_3, P_4, P_5, P_6, P_7, P_8, P_9\}$ can have Messi $n$ times, $0 \leq n \leq 4$ .

There are ${}^{(8-n)}C_n$ ways to place $n$ Messi in sequence $\{P_3, P_4, P_5, P_6, P_7, P_8, P_9\}$ . ( $^*$ Proof below).

In sequence $\{P_3, P_4, P_5, P_6, P_7, P_8, P_9, P_{10}\}$ , there are $n$ Messi , the subsequent place can have $10$ choices and the remaining $(8-2n)$ places can have $9$ choices.

$\therefore$ Total Permutations of $\{P_3, P_4, P_5, P_6, P_7, P_8, P_9, P_{10}\}$ for $n$ occurrence of Messi $= {}^{(8-n)}C_n \times 10^m \times 9^{(8-2n)}$

$\Rightarrow$ Total Permutations of $\{P_3, P_4, P_5, P_6, P_7, P_8, P_9, P_{10}\}$ $=\sum_{n=0}^{4}{}^{(8-n)}C_n \times 10^m \times 9^{(8-2n)}$

Since there are $10$ choices for $P_2$ , therefore total permutations $= 10 \times \sum_{n=0}^{4}{}^{(8-n)}C_n \times 10^m \times 9^{(8-2n)}$

$= 10 \times (1 \times 10^0 \times 9^8 + {}^7C_1 \times 10^1 \times 9^6$ $+ {}^6C_2 \times 10^2 \times 9^4 + {}^5C_3 \times 10^3 \times 9^2 + {}^4C_4 \times 10^4 \times 9^0)$

$= 909090910$

$$

$^*$ There are $^{(8-n)}C_n$ ways to place $n$ Messi in sequence $\{P_3, P_4, P_5, P_6, P_7, P_8, P_9\}$ .

Proof:

There are $(7-n)$ players to be placed between $n$ occurrences of Messi .

$\_ X \_ X \_ \ldots \_X \_ X \_$

This is equivalent to placing $n$ Messi in $(7-n + 1)$ empty slots and there are exactly ${}^{(8-n)}C_n$ ways to do so...

This problem is similar to the "Ridiculous NBA" problem posted at this site Ridiculous NBA

Please excuse the images. I could not format it as per the requirements for posting here.

Sorry i am a little short on time right now to post a detailed approach, but after constructing and working on the problem, the following generalized formula emerges.

Let $n$ be the number of passes.

Let $p$ be the number of players.

If $n$ is even, we have the number of ways ---

$\frac{{(p-1)^n +(p-1)}}{p}$

If $n$ is odd, we have the number of ways - - -

$\frac{{(p-1)^n -(p-1)}}{p}$