In how many ways can you choose three distinct numbers ....

Question

In how many ways can you choose three distinct numbers from the set of {1,2,3,...,19,20} such that their product is divisible by 4 ?

Answer 1

Strategy: There are $\binom{20}{3}$ possible choices. Let us see how many are bad.

We could choose all odd. Easy to count.

We could choose $2$ odd, and the other divisible by $2$ but not by $4$, that is, one of the numbers $2,6,10,14,18$. Again, it should not be hard to count these.

Answer 2

You can break it down like this. First we ask how many numbers are divisible by four in the set $\{1,...,20\}$ that is $4,8,12,16,20$. If exactly one is chosen from this list we have $$5\cdot {15 \choose 2}$$ ways of doing this. If two are chosen we have $${5 \choose 2} \cdot 15$$ ways of doing that. And finally there are $${5 \choose 3}$$ ways of picking three of the numbers divisible by four.

Then ask how many ways are there to choose two numbers divisible by only two? This includes $2,6,10,14,18$ and then to avoid over counting we can only choose a number not divisible by four from the rest. If we choose it to be an odd number we find: $${5 \choose 2} \cdot 10,$$ and if we choose three even numbers (not divisible by four) we find: $${5 \choose 3}$$ ways of doing this.

The sum of all of these is the answer.

Answer 3

1. We need three numbers, one of which is divisible by 4, and any other two

   In this set, the divisible by 4 are only 5 $\{4,8,12,16,20\}$. And we don't care what the other 2 will be, as they all will do the job. =>$5\cdot19\cdot18\div2 = 855$

2. We need three numbers, two of which are divisible by 2, but not by 4, and one undivisible by 4

   In the set, the divisible by 2, but not by 4, are only 5$\{2,6,10,14,18\}$. We need 2 of this set, and 1 from the set, exlcuding only divisible by 4 $\{1,2,3,5,...,19\}$, which are 15, but since we used 2 of them, there are only 13 left. => $5\cdot4\cdot13\div2=130$

Giving us a total of $985$ ways