Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?
$\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2$
There are two ways to arrange the red beads, where $R$ represents a red bead and $-$ represents a blank space.
$1. R - R - R -$
$2. R - - R - R$
In the first, there are three ways to place a bead in the first free space, two for the second free space, and one for the third, so there are $6$ arrangements.
In the second, a white bead must be placed in the third free space, so there are two possibilities for the third space, two for the second, and one for the first. That makes $4$ arrangements. There are $6+4=10$ arrangements in total. The two cases above can be reversed, so we double $10$ to get $20$ arrangements. Also, in each case, there are three ways to place the first red bead, two for the second, and one for the third, so we multiply by $6$ to get $20\cdot 6 = 120$ arrangements. There are $6! = 720$ total arrangements so the answer is $120/720 = \boxed{1/6}$.