Three numbers are chosen at random without replacement from 1, 2, 3,...,10. The probability that the minimum of the chosen numbers is 4 or their maximum is 8, is:

11/40

Total no. of ways of selecting 3 numbers = $\frac{10×9×8}{2×3}=120$ (not allowing permutation)

No. of favourable says = no. of ways to select one no. as 4 and other nos for (5 to 10)

+ no. of ways to select one no. as 8 and other nos from (1 to 7)

– no. of ways of select 4 and 8 and other from 5 –7 (as it has been counted twice)

⇒ No. of favlurable ways = $\frac{5×6}{2}+\frac{6×7}{2}-3=33$

Probability = $\frac{33}{120}=\frac{11}{40}$

(Here we didn’t allow arrangements. Even if we had, it would not have mattered as both N^r and D^r had to be multiplied by 2).