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Question

(A) The number of ways to choose $k$ non-consecutive numbers from the first $n$ natural numbers is given by the formula $\binom{n-k+1}{k}$.Here,$n = 2

Vedclass · Accepted Answer

$\frac{28}{57}$

Vedclass · Answer

(A) The number of ways to choose $k$ non-consecutive numbers from the first $n$ natural numbers is given by the formula $\binom{n-k+1}{k}$.Here,$n = 20$ and $k = 4$.Number of ways to choose four non-consecutive numbers $= \binom{20-4+1}{4} = \binom{17}{4}$.Total number of ways to choose any four numbers from $20$ is $\binom{20}{4}$.Probability $= \frac{\binom{17}{4}}{\binom{20}{4}} = \frac{\frac{17 	imes 16 	imes 15 	imes 14}{4 	imes 3 	imes 2 	imes 1}}{\frac{20 	imes 19 	imes 18 	imes 17}{4 	imes 3 	imes 2 	imes 1}} = \frac{17 	imes 16 	imes 15 	imes 14}{20 	imes 19 	imes 18 	imes 17} = \frac{16 	imes 15 	imes 14}{20 	imes 19 	imes 18} = \frac{4 	imes 15 	imes 14}{5 	imes 19 	imes 18} = \frac{4 	imes 3 	imes 14}{19 	imes 18} = \frac{12 	imes 14}{19 	imes 18} = \frac{2 	imes 14}{19 	imes 3} = \frac{28}{57}$.

Four distinct numbers are randomly selected out of the set of first $20$ natural numbers. The probability that no two of them are consecutive is -

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