Bead 7

No, the video is correct, bead 7 must be G in this scenario because any portion of 8 consecutive beads must contain at least 1 bead of each color - G O P R Y . So far we have the first six beads as follows: P Y R P Y R __ __ we only have two open slots left in this 8 bead sequence and one of the remaining two beads must be O and one must be G to comply with that rule, otherwise we only have 3 colors not 5 in this 8 bead sequence. Since O cannot be next to red, O must be #8 and G must be #7. P Y R P Y R G O __

I was tossing up between the eighth beed being Green and the Ninth beed being yellow. The chain leaves 3 beads left, which two of those need to be filled with a Green and an orange. The orange can't go in 7 because that will be next to red. Couldn't it be true if two greens were placed in 7th and 8th, and an orange ninth?

Happy to help! Let's start from the point you mentioned where we have three beads left. Our diagram looks like: P Y R P Y R __ __ __ 1 2 3 4 5 6 7 8 9 Rule #4 says that any portion of the strand with 8 consecutive beads must have at least one bead of each color. In spots 1-8, we currently only have P, Y, and R. This means we are missing G and O. We need to place G and O in spots 7-8 so that we can have at least one bead of each color within the strand of eight. Rule #3 says that O cannot be next to R, so O is placed 8th and G is placed 7th. If we place two green beads in 7 and 8 as you suggest, the strand of eight beads (1-8) would be missing the color orange and would therefore violate Rule #4. Therefore, G cannot be 8th. Does that make sense? Please let us know if you have any other questions!