I have a bag with 6 marbles numbered from 1 to 6 Mathew has a bag with 12 marbles numbered from 1 to 12 Mathew chooses one marble from his bag and I choose two from mine. In how many ways can we choose the marbles (where the order of my choices does matter) such that the sum of the numbers on my marbles equals the number on his?
+499 Let's do some casework!
First, start with Matthew picking 1. This is irrelevant to the scenario, since you can't draw two numbers and get a sum of 1, so discard that base.
Next, let's move on to 2. There's only 1 case : 1 + 1
3 : there's 2 cases, 1+2 and 2+1
4: there's 3 cases, 3+1, 1+3, 2+2
5 : there's 4 cases, 1+4, 2+3, 3+2, 4+1
6: there's 5 cases: 1+5, 2+4, 3+3, 4+2, 5+1
Clearly, there appears to be a pattern here.
for every \(n\)th case, it appears that there exist \(n-1\) ways to satisfy that case.
Assuming this is true, we then have the total number of cases as:
\((2-1) + (3-1) +(4-1)...... (12-1) = (1+2+3....11)\)
We know that the sum of the first n positive numbers is equal to \(n(n+1)/2\)(if not, I hope you read up some more about gauss and his series!).
Our answer is then:
\(11(12)/2 = 11*6 = 66\) ways
