# In how many ways can you choose three distinct numbers … [closed]

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 ?

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• It might be easier to find number of triples which are not divisible by 4 – Jack Yoon Jul 21 '14 at 14:08

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.

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.

• (I think you may have to correct for over-counting in the first step, since you could choose 4, 5, 8 or 8, 5, 4.) – user84413 Jul 21 '14 at 14:36
• You are right. I think it just needs to be divided by 2... – Joel Jul 21 '14 at 14:39
• It might be easier to break the first step into cases, depending on the number of multiples of 4 chosen. (I believe $5\binom{19}{2}$ is odd.) – user84413 Jul 21 '14 at 15:01
• I should never do combinatorics without my coffee. I think this fixes it. – Joel Jul 21 '14 at 15:13
• This looks almost right, but I think maybe there is slight over-counting in the second step. (For example, you could have 2, 6, 10 or 10, 2, 6.) – user84413 Jul 21 '14 at 15:30
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

• This is not correct. Do you see why? – Jack Yoon Jul 21 '14 at 14:20
• Yep, thanks, already updated it. – Peter Jul 21 '14 at 14:42
• Still not true. Following Andre's method I get 795. Firstly dividing by two in case 1 is not enough. – Jack Yoon Jul 21 '14 at 15:04