Can you generate the explanation of this question---Question 5. (2020 AMC Intermediate Q26, challenging) In Australian Rules football, a team scores six points for a 'goal' and one point for a 'behind'. During a game, Vladislav likes to record his team's score with a sequence of sixes and ones. There are exactly three distinct sequences which give a final score of 7 points, namely 6,1 and 1,6 and 1,1,1,1,1,1,1. How many different sequences give a final score of 20 points?

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Let's understand this Australian Rules football scoring problem. In this sport, a team scores 6 points for a goal and 1 point for a behind. We need to find how many different sequences of goals and behinds can give us exactly 20 points. The order matters - for example, scoring a goal then a behind is different from scoring a behind then a goal. Now let's set up the mathematical framework. We'll use k to represent the number of goals scored. Since each goal is worth 6 points, k goals contribute 6k points. The remaining points, which is 20 minus 6k, must come from behinds worth 1 point each. This gives us the equation 6k plus 20 minus 6k equals 20, which simplifies to 20 equals 20. Since we need 6k to be at most 20, k can be 0, 1, 2, or 3. Now let's analyze this as a combinatorial problem. For each value of k, we have a sequence of total length 20 minus 5k. We need to choose k positions out of these for our goals, and the remaining positions will have behinds. This gives us the binomial coefficient C of 20 minus 5k choose k. For k equals 0, we have 20 positions and choose 0, giving us 1 way - all behinds. For k equals 1, we have 15 positions and choose 1, giving us 15 ways to place one goal among 14 behinds. Let's complete the calculation for the remaining cases. For k equals 2, we have a sequence length of 10, and we choose 2 positions for goals out of 10, giving us C of 10 choose 2, which equals 45. For k equals 3, we have a sequence length of 5, and we choose 3 positions for goals, giving us C of 5 choose 3, which equals 10. Adding all cases together: 1 plus 15 plus 45 plus 10 equals 71. Therefore, there are 71 different sequences that give a final score of 20 points. Let's summarize our solution method. We identified that k represents the number of goals, with the remaining points coming from behinds. We found the constraint that k must be between 0 and 3. Using combinatorics, we calculated the number of ways to arrange k goals and 20 minus 6k behinds in sequences of length 20 minus 5k. Adding up all cases: 1 plus 15 plus 45 plus 10 gives us our final answer of 71 different sequences that can produce exactly 20 points in Australian Rules football.