Set A: {1, 3, 4, 6, 9, 12, 15}

If three numbers are randomly selected from set A without replacement, what is the probability that the sum of the three numbers is divisible by 3?

A. 3/14
B. 2/7
C. 9/14
D. 5/7
E. 11/14

Note: First, let’s go over the answer in shorthand notation. If this notation is unfamiliar to you, or if you would like to brush up on some fundamentals, we’ll go into details after the answer.

In all probability, we need to establish the total outcomes and the desired outcomes. How do we get only sums that are divisible by three within the constraints?

The total way to choose three items from a larger group of 7 items is 7C3

Now, what about the desired outcomes? How do we select random numbers so they sum to multiples of 3 from the set provided?

• all the sums using multiples of 3 will be divisible by 3
• the sum of 2 multiples of 3 with one non-multiple of 3 will not be divisible by 3
• the sum of 1 multiple of 3 with 2 non-multiples could work but if the sum of two non-multiples is divisible by 3

{1,3,4,6,9,12,15}
we can see that 1 and 4 are the non-multiples

There are 5C3 ways to select multiples of 3 to meet our first bullet point

What about the third possibility with 1 multiple and two non-multiples? 4+1 = 5. If we had a 1, 7, 10… then when we add them we’ll have a multiple of three, but the list lacks these choices so bullet point three is out.

Therefore desired/total: 5C3/7C3 = 2/7 will be the answer

B.

Let’s recap Permutations and Combinations. Basically, let’s select a group of things from a group of things.

First let’s discuss n! and its usage and meaning. When we want to select or re-order everything we can use factorial. Let’s arrange 4 gumdrops in a straight line on a table. How many ways can we do this? 4x3x2x1 or 4!

What about if we had 16 gumdrops and we wanted to take a group of things? Let’s say we had 16 almonds and we wanted to take 3 of them. How many groups of three are possible?

n is the total number of things and r is what we want to select.

There you go, there are 3,360 ways to select a group of three almonds from 16 almonds. Now, there’s nothing unique about each almond. Each one looks like the next. They aren’t different colors or drastically different in size. What would happen if we wanted to make a group of things from a larger group of things where each thing is unique? Something like a group of people.

Let’s take the previous idea and build on it. If I have 5 people and select three people from them I have an issue. What if I chose the same people over multiple choices but they were disguised by being in a different order. For example, one pick I chose Amy, Bill, Cathy then another pick I chose Cathy, Amy, Bill. That’s the same group but in a different order.

So when things are unique (like people are) we need to take away all the choices that have been repeated. We do that by dividing by the r! (the arrangements of the smaller group) that has repeated. It’s all done for you in the handy formula

This is like the former formula with the extra bit to divide away the times we repeated ourselves.

So in our solution, when I said 7C3, I was saying take the 7 numbers in the set and choose 3 of them but take away the times we repeated ourselves.