Question 1073434
1) There are {{{10-1=9}}} A-->X rays connecting point A to the other {{{9}}} points.
There are also {{{9}}} rays originating in each of the other points.
There is a total of {{{9*10=90}}} rays.
However, each pair of points, accounting for {{{2}}} rays, determines only {{{1}}} line.
The number of lines associated with the {{{90}}} possible rays is
{{{90/2=highlight(45)}}} .
 
2) That would be the number of lines that can be made with the other {{{8}}} points.
Using the same reasoning used for part 1, we can calculate it as
{{{8*7/2=4*7=highlight(28)}}} .
Another way to the answer:
There are {{{8}}} rays originating in point A that do not pass through B.
There are {{{8}}} rays originating in point B that do not pass through A.
Altogether, that adds to {{{8+8=16}}} rays (and lines) connecting A or B
with another point,
and none of those lines is counted twice.
Then there is {{{1}}} more line, passing through A and B: line AB.
So, there is a total of {{{8+8+1=17}}} lines passing through A and/or B.
The remaining {{{45-17=highlight(28)}}} lines .
 
3) That is how many sets of {{{3}}} points we can pick out of that set of {{{10}}} points.
Using what was taught in math class about combinations, we can calculate that number as
{{{matrix(2,1,10,3)=10!/(3!*7!)=10*9*8/2*3=10*3*4=highlight(120)}}} .
A fifth-grader could also decide that there would be
{{{10*9*7}}} possible 3-letter sequences,
but that each triangle could have its vertices listed as
{{{6}}} different sequences (such as ABC, ACB, BAC, BCA, CAB, and CBA), meaning that the number of triangles is
{{{10*9*8/6=720/6=highlight(120)}}} .
 
4) The number of triangles that contain the point A
is the number of pairs (sets of 2 items, not ordered lists of 2 items)
that can be made from the other {{{10-1=9}}} letters.
That is {{{matrix(2,1,9,2)=9!/(2!*7!)=9*8/2=9*4=highlight(36)}}} .
 
5) The number of triangles that contain side AB
is the number of points available to become the third vertex:
{{{10-2=highlight(8)}}} .
That part is easy enough for a 3-year old,
but it may be difficult if you have been conditioned to think
that you need help from formal education to answer every question.