Consider the set A = {1, 2, 3, ldots, 30}. Th | vedclass

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(A) The set $A = \{1, 2, 3, \ldots, 30\}$ contains $30$ elements.Multiples of $3$ in $A$ are $\{3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}$ (total $10$ num

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$1590$

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(A) The set $A = \{1, 2, 3, \ldots, 30\}$ contains $30$ elements.Multiples of $3$ in $A$ are $\{3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}$ (total $10$ numbers).Multiples of $9$ in $A$ are $\{9, 18, 27\}$ (total $3$ numbers).Numbers in $A$ that are not multiples of $3$ are $30 - 10 = 20$ numbers.Numbers in $A$ that are multiples of $3$ but not multiples of $9$ are $10 - 3 = 7$ numbers.To make the product divisible by $9$,we consider the following cases:Case $I$: All three numbers are multiples of $3$.The number of ways is $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.Case $II$: Exactly two numbers are multiples of $3$ and one is not.However,if we pick two multiples of $3$,their product is divisible by $9$ only if at least one of them is a multiple of $9$ $OR$ if both are multiples of $3$ (since $3 	imes 3 = 9$).Let $M_3$ be the set of multiples of $3$ $(|M_3|=10)$ and $N$ be the set of non-multiples of $3$ $(|N|=20)$.Total ways to choose $3$ numbers such that the product is divisible by $9$ is (Total ways) - (Ways where product is $NOT$ divisible by $9$).Product is $NOT$ divisible by $9$ if:$1$. No number is a multiple of $3$: $^{20}C_3 = \frac{20 	imes 19 	imes 18}{6} = 1140$.$2$. Exactly one number is a multiple of $3$ (but not $9$): $^{7}C_1 	imes ^{20}C_2 = 7 	imes 190 = 1330$.Total ways to choose $3$ numbers = $^{30}C_3 = \frac{30 	imes 29 	imes 28}{6} = 4060$.Ways divisible by $9 = 4060 - (1140 + 1330) = 4060 - 2470 = 1590$.

Consider the set $A = \{1, 2, 3, \ldots, 30\}$. The number of ways in which one can choose three distinct numbers from $A$ such that the product of the chosen numbers is divisible by $9$ is:

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