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From a class of \(\mathrm{20}\) students whose names are listed in alphabetical order, a teacher will choose one group of \(\mathrm{3}\) students to represent the class in a student congress. If the teacher will not choose a group of \(\mathrm{3}\) students whose names are in \(\mathrm{3}\) consecutive positions on the list, how many different groups of \(\mathrm{3}\) students could be chosen by the teacher?
Let's break down what this problem is asking in plain English. We have 20 students whose names are arranged alphabetically on a list - think of it like positions 1, 2, 3, ..., 20. The teacher wants to pick 3 students to represent the class, but there's one important restriction: she cannot pick 3 students whose names appear right next to each other on the alphabetical list.
For example, if the students in positions 5, 6, and 7 were selected, that would be forbidden because their names are consecutive on the alphabetical list. However, selecting students from positions 3, 8, and 15 would be perfectly fine since they're not consecutive.
Our goal is to find how many different groups of 3 students can be chosen while respecting this restriction.
Process Skill: TRANSLATE - Converting the problem's English description into clear mathematical requirements
Let's start by imagining there were no restrictions at all. How many ways can we choose 3 students from a class of 20?
Think of it this way: we're picking 3 students where the order doesn't matter (student A, B, C is the same group as student C, A, B). This is a combination problem.
We can calculate this step by step:
- For the first student, we have 20 choices
- For the second student, we have 19 remaining choices
- For the third student, we have 18 remaining choices
This gives us \(20 \times 19 \times 18 = 6,840\) ways. But wait - this counts each group multiple times because we're treating the order as important (picking A then B then C is different from picking C then A then B).
Since we have 3 students in each group, each group gets counted \(3! = 6\) times. So the actual number of unique groups is:
\(6,840 \div 6 = 1,140\)
Using combination notation: \(\mathrm{C(20,3)} = \frac{20!}{3! \times 17!} = 1,140\)
Now let's count how many groups are forbidden - those where all 3 students have consecutive positions on the alphabetical list.
What does "3 consecutive positions" mean? It means selecting students from positions like (1,2,3) or (5,6,7) or (18,19,20).
Let's think systematically: if we pick 3 consecutive students starting from position k, we select students in positions k, k+1, and k+2.
What values can k take?
- If k = 1, we select positions 1, 2, 3 ✓
- If k = 2, we select positions 2, 3, 4 ✓
- If k = 3, we select positions 3, 4, 5 ✓
- ...
- If k = 18, we select positions 18, 19, 20 ✓
- If k = 19, we would select positions 19, 20, 21 ✗ (position 21 doesn't exist)
So k can range from 1 to 18, giving us exactly 18 forbidden groups.
Now we can use complementary counting, which follows this simple logic:
Number of valid groups = Total possible groups - Forbidden groups
We calculated:
- Total possible groups (without restrictions) = 1,140
- Forbidden groups (3 consecutive students) = 18
Therefore:
Number of valid groups = \(1,140 - 18 = 1,122\)
Process Skill: APPLY CONSTRAINTS - Using the restriction to eliminate invalid cases from our total count
The teacher can choose 1,122 different groups of 3 students that satisfy the given restriction.
Looking at our answer choices, this matches option (B) 1122.
Answer: (B) 1122
1. Misinterpreting what "consecutive positions" means
Students often confuse "consecutive positions on the alphabetical list" with "any three students whose names start with consecutive letters." The problem refers to positions 1, 2, 3 or 5, 6, 7 on the ordered list, not students named like "Alice, Bob, Charlie" whose names happen to start with A, B, C. This misunderstanding leads to completely wrong counting approaches.
2. Attempting direct counting instead of complementary counting
Many students try to directly count valid groups by considering complex cases like "at least one gap between selected positions." This approach quickly becomes overwhelming with multiple subcases. The complementary counting approach (total groups minus forbidden groups) is much simpler, but students often don't recognize this strategic advantage.
3. Confusing combinations with permutations
Since we're selecting groups where order doesn't matter, this is a combinations problem \(\mathrm{C(20,3)}\). However, students sometimes think the specific positions matter and try to use permutations \(\mathrm{P(20,3)}\) instead, leading to a much larger and incorrect starting count.
1. Miscounting the forbidden consecutive groups
When counting groups of 3 consecutive positions, students often incorrectly count 20 forbidden groups instead of 18. They forget that if you start at position 19, you'd need positions 19, 20, 21 - but position 21 doesn't exist. The valid starting positions for consecutive triplets are only 1 through 18.
2. Arithmetic errors in combination calculation
When calculating \(\mathrm{C(20,3)} = \frac{20!}{3! \times 17!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1}\), students frequently make computational mistakes. Common errors include: forgetting to divide by \(3! = 6\), miscalculating \(20 \times 19 \times 18 = 6,840\), or making division errors when computing \(6,840 \div 6 = 1,140\).
1. Selecting the total count instead of the restricted count
After correctly calculating that total possible groups = 1,140 and forbidden groups = 18, some students accidentally select 1,140 as their final answer. They forget to complete the final subtraction step \((1,140 - 18 = 1,122)\) required by the complementary counting approach, especially if 1,140 appears among the answer choices.