Question 611286
Note: I'm assuming that all 6 flags are different and that order matters.

a)


Since A and B must be next to each other AND A must be to the left of B, this means that we can effectively "combine" the two flags to make a third flag (say AB). This is not a single flag, but because the two can't be separated or ordered differently, we can group the two like this.


So we have the 5 flags:  AB, C, D, E, F


Where AB is considered one flag.


From here, there are 5! = 5*4*3*2*1 = 120 different ways to order these five flags.


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b)


This is a lot like part A, but now we consider the case that we're dealing with the "flag" BA". This will double the number in part a) to give you 2*120 = 240 



Let me know if this makes sense. Thanks.