+305 Not including the identity transformation, the eleven transformations that preserve the regular hexagon shown are counterclockwise rotations by \(60, 120, 180, 240, 360\) and reflections across the six dashed lines shown. Kristina randomly picks six transformations \(T_1, T_2, T_3, T_4, T_5\)and \(T_6\) with replacement, from this set of eleven. She performs these six transformations on the hexagon, in succession. The probability that the point \(P\) is transformed to each of the hexagon's six vertices exactly once during this process is \(\frac{k}{11^6}\) What is the value of \(k\)?
Ok so at first I thought i was 720 because 6!, but there's multiple ways to get to the next vertex, and its different for each vertex, so I don't know what to do.
My thought process is since it gives us the denominator, its just asking how many ways P can touch all the vertexes of the hexagon through choosing 6 transformations. Also, when you make a transformation, I think you can remove the line that reflects that point back onto itself. But I got lost on how to count the counterclockwise rotations.
Thx! - (^-^)
+905 To solve this problem, we need to consider the transformations that can move point P to each vertex exactly once.
The value of k in the probability $\frac{k}{11^6}$ can be calculated as follows:
There are 6! = 720 ways to arrange the six vertices in a sequence.
For each sequence, we need to count the number of ways to choose transformations that achieve that sequence:
The first transformation can be any of the 11 that moves P to the first vertex in the sequence.
The second transformation must be one of the 2 that moves the current position to the second vertex.
Each subsequent transformation also has 2 choices to move to the next vertex.
Therefore, for each sequence, there are 11 * 2^5 = 352 ways to choose the transformations.
The total number of favorable outcomes is thus 720 * 352 = 253,440.
Therefore, k = 253,440.
+305 Ok I did the problem again and I disagree. Here's my approach.
For the first transformation, any of the 11 options will achieve a satisfactory result (none of the 6 points has yet been transformed to). After that, the remaining 5 points can be transformed to in any order, as long as there is no repetition, so 5! = 120 orders for visiting points. At each step after the first, there are two transformations (one rotation and one reflection) that will get from one point to the next [distinct] point. Thus, the overall probability of success is
\(120\times\frac{11}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11}\times\frac{2}{11} = \frac{42,240}{11^6}\)
so k = 42240
EDIT -------
I think you did 6! instead of 5!
Since you counted that the first transformation can be any of the 11 moves, then there are only 5! ways to arrange the other 5
