For the next three (03) items that follow : Five numbers are randomly picked from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and arranged them in ascending order [\(x_1 < x_2 < x_3 < x_4 < x_5\); \(x_1\) being the smallest number and \(x_5\) being the largest number selected].
What is \(P(x_1 = 2, x_2 = 3, x_3 = 8)\) equal to?
\(\dfrac{1}{252}\)
For \(x_1=2, x_2=3, x_3=8\), the number 1 cannot be chosen (else \(x_1=1\)) and none of 4, 5, 6, 7 can be chosen (else \(x_3\) would be smaller than 8). Since only two numbers, 9 and 10, remain greater than 8, they must both be chosen as \(x_4\) and \(x_5\). So the only favourable set is \(\{2,3,8,9,10\}\), giving \(P = \dfrac{1}{\binom{10}{5}} = \dfrac{1}{252}\).
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