Loading...
A folk group wants to have one concert on each of the seven consecutive nights starting January 1 of next year. One concert is to be held in each of cities A, B, C, D and E. Two concerts are to be held in city F, but not on consecutive nights. In how many ways can the group decide on the venues for these seven concerts?
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:
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
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
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).
\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
Now we combine our results using the multiplication principle:
Total arrangements = \(15 \times 5!\)
Looking at our answer choices, this matches option (C) \(15 \times 5!\)
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.
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.
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 \(\mathrm{C(6,2)} = 15\).
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.