A folk group wants to have one concert on each of the seven consecutive nights starting January 1 of next...

1. Translate the problem requirements: We have 7 consecutive nights, need to place concerts in cities A, B, C, D, E (one each) and city F (exactly two, but not consecutive). Find total arrangements.
2. Identify valid positions for the two F concerts: Since F concerts cannot be consecutive, determine how many ways we can choose 2 non-consecutive positions from 7 nights.
3. Arrange the remaining concerts: Once F positions are fixed, arrange the 5 distinct concerts (A, B, C, D, E) in the remaining 5 positions.
4. Apply multiplication principle: Multiply the number of ways to place F concerts by the number of ways to arrange the other concerts.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we need to figure out. We have 7 consecutive nights (January 1st through January 7th). We need to schedule concerts in different cities with these specific rules:

• Cities A, B, C, D, and E each get exactly one concert
• City F gets exactly two concerts, but these two F concerts cannot happen on back-to-back nights

We want to count how many different ways we can assign these concerts to the 7 nights.

Process Skill: TRANSLATE - Converting the constraint language into clear mathematical requirements

2. Identify valid positions for the two F concerts

Since the two F concerts cannot be on consecutive nights, let's think about this systematically. We need to choose 2 positions out of 7 nights such that they are not next to each other.

Let's imagine the 7 nights as positions: 1, 2, 3, 4, 5, 6, 7

One way to think about this: if we place the two F concerts in non-consecutive positions, we can think of it as placing the 5 other concerts (A, B, C, D, E) first, and then inserting the 2 F concerts into the gaps.

When we place 5 concerts in a row, we create 6 possible gaps (before the first, between each pair, and after the last):
_X_X_X_X_X_

We need to choose 2 of these 6 gaps for our F concerts. The number of ways to choose 2 gaps from 6 gaps is \(\mathrm{C(6,2)}\).

\(\mathrm{C(6,2)} = \frac{6!}{2! \times 4!} = \frac{6 \times 5}{2 \times 1} = 15\)

Process Skill: VISUALIZE - Using gap-insertion method to handle the non-consecutive constraint

3. Arrange the remaining concerts

Once we've decided where the two F concerts will go, we have 5 remaining positions that need to be filled with concerts for cities A, B, C, D, and E.

Since these are 5 different cities, we need to arrange 5 distinct objects in 5 positions. This is simply \(5!\) (5 factorial).

\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

4. Apply multiplication principle

Now we combine our results using the multiplication principle:

• Number of ways to place the 2 F concerts in non-consecutive positions: 15
• Number of ways to arrange the remaining 5 concerts: \(5!\)

Total arrangements = \(15 \times 5!\)

Looking at our answer choices, this matches option (C) \(15 \times 5!\)

Final Answer

The answer is (C) \(15 \times 5!\)

Our solution shows that there are 15 ways to position the two F concerts such that they don't occur on consecutive nights, and for each such positioning, there are \(5!\) ways to arrange the concerts for cities A, B, C, D, and E in the remaining nights.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the non-consecutive constraint
Students often misread "not on consecutive nights" as "must be separated by at least 2 nights" or confuse it with other spacing requirements. This leads them to incorrectly calculate the valid positions for the F concerts.

2. Attempting direct counting instead of using systematic methods
Many students try to list out all possible arrangements manually rather than recognizing this as a gap-insertion problem. This approach becomes unwieldy with 7 positions and often leads to missed cases or double-counting.

3. Failing to separate the positioning and arrangement sub-problems
Students may not recognize that this problem breaks into two independent parts: (1) choosing positions for the F concerts with constraints, and (2) arranging the remaining concerts. Instead, they try to solve everything simultaneously, making the problem much more complex.

Errors while executing the approach

1. Incorrectly calculating \(\mathrm{C(6,2)}\)
When using the gap-insertion method, students may make arithmetic errors in computing \(\mathrm{C(6,2)} = \frac{6!}{2! \times 4!} = 15\). Common mistakes include forgetting to divide by 2! or miscalculating the basic combination formula.

2. Confusing the number of gaps created
Students may incorrectly think that 5 concerts create 5 gaps instead of 6 gaps. This fundamental error in visualizing the gap-insertion method leads to calculating \(\mathrm{C(5,2)} = 10\) instead of the correct \(\mathrm{C(6,2)} = 15\).

Errors while selecting the answer

1. Choosing the wrong coefficient with \(5!\)
Students who make errors in the positioning calculation might arrive at \(10 \times 5!\) (choice A) if they miscounted gaps, or \(14 \times 5!\) (choice B) through other calculation mistakes, but still correctly identify that \(5!\) represents the arrangement of the remaining concerts.