Question 1112702
<font face="Times New Roman" size="+2">

Presuming a single file line and presuming that reversing the position of two sign carriers or two pom-pom carriers does not result in a different arrangement, then, since two of the sign carriers are accounted for, the only variability is the position of the one remaining sign carrier relative to the four pom-pom carriers.  The remaining sign carrier could be in any one of the five positions occupied by that sign carrier and the four pom-pom carriers.  Hence, five different arrangements.


On the other hand, if you are considering that an arrangement where Lucy is the sign carrier in the front of the line, Alice is the sign carrier in the back of the line, and Susan is the sign carrier in the exact middle as a different arrangement from Susan in front, Lucy in the back, and Alice in the middle, then you have an entirely different calculation.  I'm going to assume that this is NOT what you meant.  I only mention this to illustrate that your problem, as stated, is ambiguous and it is your responsibility to make sure you communicate clearly if you want to get correct answers.  


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>