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(C) Total ways to choose $4$ numbers from $20$ is $n(S) = {}^{20}C_4 = \frac{20 	imes 19 	imes 18 	imes 17}{4 	imes 3 	imes 2 	imes 1} = 4845$.L

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Statement-$1$ is true,Statement-$2$ is false.

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(C) Total ways to choose $4$ numbers from $20$ is $n(S) = {}^{20}C_4 = \frac{20 	imes 19 	imes 18 	imes 17}{4 	imes 3 	imes 2 	imes 1} = 4845$.Let the $A.P.$ be $a, a+d, a+2d, a+3d$. Since $1 \le a$ and $a+3d \le 20$,we have $3d \le 20-a \le 19$,so $d \le 6$.For a fixed $d$,the number of $A.P.s$ is $20-3d$.For $d=1$: $20-3(1) = 17$.For $d=2$: $20-3(2) = 14$.For $d=3$: $20-3(3) = 11$.For $d=4$: $20-3(4) = 8$.For $d=5$: $20-3(5) = 5$.For $d=6$: $20-3(6) = 2$.Total $A.P.s = 17+14+11+8+5+2 = 57$.Probability $= \frac{57}{4845} = \frac{1}{85}$.Statement-$1$ is true.Statement-$2$ claims the common difference $d$ can only be $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5$. However,$d=6$ is also possible (e.g.,$1, 7, 13, 19$). Thus,Statement-$2$ is false.

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