SOLUTION: Solve the following:
a. On a piece of paper, 8 points are marked such that no 3 points lie on the same straight line. How many lines can be drawn passing through any 2 of these
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-> SOLUTION: Solve the following:
a. On a piece of paper, 8 points are marked such that no 3 points lie on the same straight line. How many lines can be drawn passing through any 2 of these
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Question 1172246: Solve the following:
a. On a piece of paper, 8 points are marked such that no 3 points lie on the same straight line. How many lines can be drawn passing through any 2 of these points
b. Seven points lie on a circle. How many triangles can be drawn using any 3 of these points as vertices?
c. A group of 4 adults and 3 children are to be formed from 8 adults and 5 children. How many possible groups are there?
d. To promote reading, a teacher decides to feature 3 classics, 4 contemporary novels and 2 non-fiction books on a notice board. How many selections can she make from 5 classics, 6 contemporary novels and 4 non-fiction books? Found 2 solutions by math_tutor2020, ikleyn:Answer by math_tutor2020(3817) (Show Source):
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.
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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
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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.
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