Problem Solving

winnerhere wrote:Three numbers are selected from first 9 natural numbers.What is the probability that the product of those numbers are divisible by 10?

1) 3/4

2) 1/12

3) 11/42

4) 5/12

5) 7/18

If the product of three numbers between 1 and 9 inclusive will be a multiple of 10, we'll need to have prime factors of 2 and 5 in our product. That is, we'll certainly need to pick the '5', and we'll also need to pick at least one even number.

We have 9C3 = (9*8*7)/3! = 84 ways of choosing three numbers between 1 and 9, inclusive (the wording of the question is bad - I'm assuming the numbers need to be different). That will be the denominator of our probability.

Now, we certainly need to pick the '5' in order to get a product which is a multiple of 10. If we pick the 5, we then have 8C2 = (8*7)/2 = 28 ways to pick the other two numbers.

We aren't done, since this includes combinations like {1,5,7} which do not contain any even numbers. So there will be less than 28 ways to pick our three numbers (and notice that we now know our probability must be less than 28/84 = 1/3 -- there are only two answer choices which are possible, B and C, and B is far too small, so we could choose C and move on). We can count all of the selections that give us a '5' along with two odd numbers -- that is, we can count the selections we do *not* want -- and subtract this from 28; that will give us the number of ways to select the '5' and at least one even number. After we pick the '5', we can select two odd numbers from the four remaining odd numbers in 4C2 = 6 ways. So we can make 28-6 = 22 selections which include the '5' and at least one even number. Dividing by 84, the total number of possible selections, gives us the answer: 22/84 = 11/42.

winnerhere wrote:My approach:

The total number of combinations possible = 9c3 = 84

For a number to be divisible by 10, it should have 5 as a factor = so one number is five

The other number should be an even number . it could be 2,4,6 or 8

The third number could be any of the 7 numbers that are left out after picking the first 2

Therefore 1 * 4 * 7 = 28

Probablity = 28/84 - but Im wrong in my approach.Please help me understand where I am lagging behind in my approach

This approach is almost right - it can be fixed to get the right answer - but the problem here is that you are double-counting. Picking the '5' is fine, but when you pick the remaining two numbers, you are counting some possibilities twice. For example, when you first pick the even number, you might pick the '2'. Then when you pick the last number, you might pick a '4'. With the method you're using, you'd count that as though it were a different selection from picking the '4' as your even number and the '2' as the last number. But, since order does not matter here (since you used 9C3 in your denominator, you have to count the numerator as though order does not matter as well), then picking 2, then 4 gives the same selection as picking 4, then 2.

So your count is slightly too high, because you've counted every pair of even numbers you might select *twice* instead of once. If you subtract 4C2 = 6, which is the number of pairs of even numbers you can select, you'd correct the double-counting error and get the right answer.