consider the set A={1,2,3,...,30}.the number if ways in which one can choose three distinct numbers from A so that the product of the choosen numbers is divisible by 9 is
Answers
Nos in set divisible by 3(but not by 9)-3,6,12,15,21,24,30-Total 7
Nos in set divisible by 9= 9,18,27-Total 3
Now the question is asked for three distinct numbers,
Therefore, the ans should be
7*6*20/2+7*6*5/6+7*6*3/2+7*3*2/2+3*2*1/6+3*20*19/2+3*2*20/2+3*7*20
=1569
Answer:
1590
Step-by-step explanation:
We solve the opposite: let Y 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 10 multiples of 3 among those 3 are also a multiple of 9 in {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 7 different ways and the others are chosen from the rest of the set providing 7×20×19 different cases. The 2nd case also provides 20×19×18 different cases and choosing three arbitrary number is possible to 30×29×28 cases. So:
X=30×29×28−20×19×18−7×20×19=14860
If we don't mind ordering the answer would then be:
X=(30.3)−(20.3)−(71)(202)=1590