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Please see the question below, my answer, and the intuition I'm leaning towards. I'd like to know if the way I'm framing this in my mind is correct but specifically if there is a more intuitive way to think about $\binom{8}{3}$ as being the number of ways to arrange the balls such that no $2$ red balls are adjacent.

$3$ red balls and $7$ blue balls are randomly placed in a row. What is the probability that no two red balls are adjacent?

Answer: $\frac{7}{15}$

Or in expressed in terms of combinations

$$\frac{\binom{8}{3}}{\binom{10}{3}}$$

Can someone explain to me the intuition of how $\binom{8}{3}$ counts the number of ways that no two red balls are adjacent? I'm imagining it's because we are "pre-selecting" two blue balls to serve as dividers, thus reducing our total to $8$. Is this correct and is there a better way to think about this intuitively?

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Imagine laying out the 7 blue balls. This divides your space into 8 areas. See below the areas are represented by "a" and the 7 blue balls by "*"

$$a*a*a*a*a*a*a*a$$

So, what's left is to pick 3 of those 8 areas to place a red ball in. $8\choose3$

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The idea here is to line up the seven blue balls in a row, which creates eight spaces into which we can place a red ball, six between successive blue balls and two at the ends of the row. By choosing three of these eight spaces for the red balls, we ensure that no two of the red balls are consecutive since at least one blue ball must lie between any two red balls. There are $\binom{8}{3}$ ways to select three of the eight spaces for the red balls.

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