Question

Consider the set a = { 1,2,3,4,......,....,30} the number of ways one can choose three numbers from the set such that the product of them is didvisible by 9

Answer 1

We solve the opposite: let YY denote the number of ways one can choose three distinct numbers for which their multiple not divisible by 9 i.e. divisible by 3 and not 9 or not divisible by 3 at all. First note that there are 1010 multiples of 3 among those 33are also a multiple of 9 in {1,2,...,30}{1,2,...,30}so the number of ways of choosing 3 numbers in case 1 is to choose exactly one multiple of 3 not of 9 to 77 different ways and the others are chosen from the rest of the set providing 7×20×197×20×19different cases. The 2nd case also provides 20×19×1820×19×18 different cases and choosing three arbitrary number is possible to 30×29×2830×29×28 cases. So:

X=30×29×28−20×19×18−7×20×19=14860X=30×29×28−20×19×18−7×20×19=14860

If we don't mind ordering the answer would then be:

X=(303)−(203)−(71)(202)=1590X=(303)−(203)−(71)(202)=1590

I hope this is correct and if it helped you please mark it brainliest