2019 Mock AMC 10B Problems/Problem 23

To solve this problem, we can consider each unit of $1$ to be separated into $n$ equal increments. (For example, if each unit of $1$ is separated into $2019$ equal increments, there would be $2019$ tiny fragments of $\frac{1}{2019}$ per unit of $1$ .) Recall that $\text{probability} = \frac{\text{favorable}}{\text{total}}$ , so we have to calculate (1) the number of ways to arrange $4$ “blocks” of $1$ in the whole “grid” of length $10$ given $n$ increments per unit (this is because the maximum least value for one of the sets is $9$ ) such that no two of them overlap and (2) the number of ways to arrange $4$ “blocks” of $1$ in the whole “grid” of length $10$ given $n$ increments per unit under no restrictions. We can easily see that the probability is

$\frac{4! \times \dbinom{6x+4}{4}}{(9x+1)^4}$ .

(The $4!$ is included due to the four blocks $S_1, S_2, S_3, S_4$ being able to be arranged order.)

Since there are infinite real numbers in a unit of $1$ , there are infinite increments in a unit of $1$ , so we should take the limit as $n$ approaches infinity:

$\text{probability} = \lim_{x \to \infty} \frac{4! \times \dbinom{6x+4}{4}}{(9x+1)^4}$ Dividing leading coefficients, we get the probability to be $\frac{6^4}{9^4}$ , which simplifies to $\frac{2^4}{3^4}$ .

Thus, our answer is $16+81=\boxed{97}$ .

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2019 Mock AMC 10B Problems/Problem 23