Question 1172246
<font color=black size=3>I'll do part (a) to get you started.


We have n = 8 points and we want to select r = 2 points from that list to generate a line.


There are nCr = 8C2 = 28 different ways to do this. The order of the two points selected doesn't matter. Segment AB is the same as segment BA.


--------------------


Here are the steps on how to compute 8C2 through the nCr combination formula
Before we do that though, let's note that 
8! = 8*7*6*5*4*3*2*1
6! = 6*5*4*3*2*1
Both right hand sides contain "6*5*4*3*2*1" which allows us to say 
8! = 8*7*6!


Now we can compute the nCr value
nCr = (n!)/(r!*(n-r)!)
8C2 = (8!)/(2!*(8-2)!)
8C2 = (8!)/(2!*6!)
8C2 = (8*7*6!)/(2!*6!) .... use what was discussed above
8C2 = (8*7)/(2!) .... the "6!" terms cancel
8C2 = 56/(2*1)
8C2 = 56/2
8C2 = 28


--------------------


Put another way:
We have 2 slots to fill and 8 items to pick from. The first slot has 8 choices and the second slot has 7 choices since we can't reuse whatever was picked in the first slot. 
This gives 8*7 = 56 permutations where order matters (eg: AB would be different from BA)
However, naming a segment can be done in two different ways. So AB is not different from BA, and order doesn't matter. We have to divide by 2 to correct for the fact we double-counted
That's why we say 56/2 = 28.
There are 28 combinations possible.
So there are 28 different lines possible.


--------------------


<font color=red>Answer: 28</font></font>