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

We define the following subsets of set A : $\mathrm{X}=9,18,27 \quad$ (divisible by 9) Number of elements $=3$ $\mathrm{Y}=3,6,12,15,21,24,30 \quad \quad($ divisible by 3 but not by 9 ) Numher if elements $=7$ $\mathrm{Z}=$ all the remaining elements of set A (not divisible by 3 ) Number of elements $=20$ Number of ways of choosing 3 distinct numbers such that their product is NOT divisible by 9 : (I) Choose any 3 numbers from subset $\mathrm{Z}={ }^{20} \mathrm{C}_{3}=1140$ (II) Choose any 2 numbers from subset $\mathrm{Z}$ and any 1 number from subset $\mathrm{Y}=$ ${ }^{20} \mathrm{C}_{2} \times{ }^{7} \mathrm{C}_{1}=1330$ Total number of ways of choosing 3 distinct numbers such that their product is NOT divisible by $9=1140+1330=2470$ Total number of ways of choosing 3 distinct numbers from set $\mathrm{A}={ }^{30} \mathrm{C}_{3}=$ 4060 So, Number of ways of choosing 3 distinct numbers such that their product is divisible by $9=4060-2470=1590$