The number of ways in which we can select thr | vedclass

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Question

(B) The total number of ways to select $3$ numbers from $30$ is given by $^{30}C_3$.$^{30}C_3 = \frac{30 	imes 29 	imes 28}{3 	imes 2 	imes 1} = 5

Vedclass · Accepted Answer

$3605$

Vedclass · Answer

(B) The total number of ways to select $3$ numbers from $30$ is given by $^{30}C_3$.$^{30}C_3 = \frac{30 	imes 29 	imes 28}{3 	imes 2 	imes 1} = 5 	imes 29 	imes 28 = 4060$.In the range $1$ to $30$,there are $15$ even numbers and $15$ odd numbers.The number of ways to select $3$ even numbers from the $15$ available even numbers is $^{15}C_3$.$^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 5 	imes 7 	imes 13 = 455$.The number of ways to select $3$ numbers such that not all of them are even is the total number of ways minus the number of ways to select $3$ even numbers.Required number of ways $= ^{30}C_3 - ^{15}C_3 = 4060 - 455 = 3605$.

The number of ways in which we can select three numbers from $1$ to $30$ so as to exclude every selection of all even numbers is

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