Consider the set $A = 1, 2, 3, . . ., 30$ . 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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Solution

Set $A = 1, 2, 3, . . ., 30$
Case $1$ :
one number from $(9, 18, 27)$
$^{3} {C_{1}}^{29} C_{2}$
Case $2$ :
two number from $(3, 6, 12, 15, 21, 24, 30)$
$^{7} {C_{2}}^{20} C_{1}$
Total number of ways $= 3 \times \frac{29 \times 28}{2} + \frac{7 \times 6}{2} \times 20$
$\begin{matrix} = 1218 + 420 = 1638 \end{matrix}$

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Consider the set A=1,2,3,...,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

Set A=1,2,3,...,30Case 1:one number from (9,18,27)3C129C2Case 2:two number from (3,6,12,15,21,24,30)7C220C1Total number of ways=3×29×282+7×62×20=1218+420=1638

Consider the set $A = 1, 2, 3, . . ., 30$ . 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