Five numbers $$x_1, x_2, x_3, x_4, x_5$$ are randomly selected from the numbers $$1, 2, 3, \ldots, 18$$ and are arranged in the increasing order $$(x_1 < x_2 < x_1 < x_4 < x_2)$$. The probability that $$x_2 = 7$$ and $$x_4 = 11$$ is

Total number of ways to select $$5$$ numbers from $$\{1,2,3,\ldots,18\}$$ is $${18\choose5}=8568$$

Given,

$$x_2=7,\qquad x_4=11$$

Since

$$x_1<x_2<x_3<x_4<x_5$$

we must have:

$$x_1\in\{1,2,3,4,5,6\}$$

So, number of choices for $$x_1$$ is

$$6$$

Also,

$$x_3\in\{8,9,10\}$$

So, number of choices for $$x_3$$ is

$$3$$

Further,

$$x_5\in\{12,13,14,15,16,17,18\}$$

So, number of choices for $$x_5$$ is

$$7$$

Hence, favourable outcomes are

$$6\times3\times7=126$$

Therefore, required probability is

$$\frac{126}{8568}=\frac1{68}$$

Hence,

$$\boxed{\frac1{68}}$$