First, we need to make a visual. I can do this by making a list, like in a data set. Let b = blue and r = red.
(b, b, b, r, r, r, r)
Note that this is not necessarily the order.
Now we can modify the order.
1. The first car is blue
This is already done!
2. The last car is blue.
We can trade the second blue with the last red.
(b, r, b, r, r, r, b)
3. The middle car is blue.
We can trade the middle red with the third blue.
(b, r, r, b, r, r, b)
This arrangement we have created follows the rules. It is one possible queue!
5. But we aren't done yet! We need to figure out if there are any more arrangements.
There is none. We know this because there are only three cars that are blue, and three spots these cars MUST park in.
Therefore, we could not swap any colors without maintaing the rules.
Answer: There is one possible arrangement of the queue:
(b, r, r, b, r, r, b)
Please let me know if this was helpful, or if you have any questions. If you have time, I'd be so grateful for a rating!
Answer:
144
Step-by-step explanation:
According to the question, we can represent the pattern of the red and blue cars in the following way below:
B, R, R, B, R, R, B
We are given that there are 3 blue and 4 red cars in total
The first blue car has 3 possibilities, and each time a blue car is parked, we now have one less possibility (the same can be done with the red cars)
So, the number of possibilities for each spot in the pattern can be multiplied together:
3 x 4 x 3 x 2 x 2 x 1 x 1 = 144
Hope this helps!
