How I did it

Follow my train of thought here:
I forgot how to use grid method and I didn't see how it would work, so I listed it all out by hand.
a+b has to be less than or equal to c to make this work, so (the addition means a+b)
1 Nothing
2 1+1
3 1+2 1+1
4 2+2 1+2 1+1
5 2+3 2+2 1+2 1+1
6 3+3 2+3 2+2 1+2 1+1
7 3+4 3+3 2+3 2+2 1+2 1+1Notice how it all stacks up? Problem is, my answer was 21 with this method, and there was no choice of 21. I didn't know what else to do, so I picked (spoiler alert!!!)
22
And I got it right lol. Was there something wrong with my method though?

Note: some of them may be in the wrong order but you can just flip it around

Another note: I thought about it over and a lot of it doesn't make sense. Still if you can correct this strategy so it works thanks in advance.

@TheDarkinBlade Hi there! I'll start by giving the original Your Turn question again just to make it easier to follow along: We want \(3\) integers \(a < b < c \) between \(1\) and \(7\) inclusive (which means that we are allowed to set them equal to \(1\) or \(7,\) such that these \(a, b,\) and \(c\) cannot be the sides of a triangle.
Note that none of the \(a, b\) or \(c\) are allowed to be equal to each other, so that means that some of the ways in your chart above aren't valid. For example, these aren't valid:So all of the ways like \(1 + 1, 2 + 2, 3 + 3 \ldots \) aren't possible, since here \(a = b.\)
Another thing I noticed is that you seem to be carrying over the previous line to the next, while tacking on one more sum. Like, for the line \(c = 3,\) you copy the \(1 + 2; 1 + 1\) over to the next line, while adding a \(2 + 2\) since the new sum can be \(4.\) However, notice that in the next line, we actually should add to the ways which have the form \( 1 + \text{ anything }.\) For example, the line \(c = 4\) should also include \(1 + 3,\) even though \(1 + 3\) doesn't occur in the previous line where \(c = 3.\)
So, a complete list of all the ways would look more like this:
$$\begin{aligned} c &=1 \implies \text{ Nothing } \\ c &= 2 \implies \text{ Nothing } \\ c &= 3 \implies 1 + 2 \\ c &= 4 \implies 1 + 2; \text{ } 1 + 3 \\ c &= 5 \implies 1 + 2; \text{ } 1 + 3; \text{ } \textcolor{red}{1 + 4}; \text{ } 2 + 3 \\ c &= 6 \implies 1 + 2; \text{ } 1 + 3; \text{ } 1 + 4; \text{ } \textcolor{red}{1 + 5}; \text{ } 2 + 3; \text{ } 2 + 4 \\ c &= 7 \implies 1 + 2; \text{ } 1 + 3; \text{ } 1 + 4; \text{ } 1 + 5; \textcolor{red}{1 + 6}; \text{ } 2 + 3; \text{ } 2 + 4; \text{ } \textcolor{red}{2 + 5}; \text{ } 3 + 4 \\ \end{aligned} $$I've highlighted some of these entries in \(\textcolor{red}{\text{red}}\) since they are additions to the pattern that we borrow from the previous line.
Now, if we add up all of these ways, we have \(1\) on the third row, \(2\) on the fourth row, \(4\) on the fifth row, \(6\) on the sixth row, and \(9\) on the seventh row, which adds up to \( 1 + 2 + 4 + 6 + 9 = \boxed{22}.\)
I hope this helps! Let me know if anything I said was confusing. Good luck with the rest of the course; you're almost done!

@TheRogueBlade Your idea is correct but you missed some pairs and also, a<b so pairs with the same numbers are incorrect. When prof. loh drew a line across the chart he did it above the middle, so he excluded all the pairs with the same numbers.
These shouldn't be included️
2: 1,1
3: 1,1
4: 2,2 1,1
5: 2,2 1,1
6: 3,3 2,2 1,1
7: 3,3 2,2 1,1I really liked the way you did the question and it would be quite useful if I didn't want to draw a graph