A Circular permutation problem

Question

i've this problem that's making me crazy. I've 12 ball, 4 of these are red, the others are white. What's the probability to obtain a circular sequences of balls in which red balls are not adjacent?

thanks a lot

Answer 1

We may assume without loss of generality that the first red ball is placed in the "top" position. Then there are $11\choose 3$ ways to arrange the remaining balls. To count the good arrangements, merge each red ball with the white ball following it; we conclude that there are $7\choose 3$ ways to complete the arrangement. The probability is therefore $$ \frac{7\choose 3}{11\choose 3}$$

Answer 2

Cut the circle so that we have a red immediately to our left, and straighten it out into a line like this: $$ \ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad R$$ There are $\binom{11}{3}$ equally likely ways to choose where the rest of the red balls will go. Now we count the number of ways in which no two red are adjacent in the original circular configuration. We must place a white at the left end, and a white next to the red at the right end, so we have reached $$ W\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad\ast\quad W\quad R$$

Now we have a "line" problem. We must choose $3$ positions from $9$ in a line to put $3$ red.

Write down $6$ occurrences of $\times$, to represent the ultimate positions of the $6$ white: $$\times\quad\times\quad\times\quad\times\quad\times\quad\times$$ These determine $7$ "gaps" ($5$ real gaps, and the two "endgaps." We must choose $3$ of these to slip an $R$ into. There are $\binom{7}{3}$ ways to do this.