Video Transcript
Given 87 points arranged in a
circle, determine the number of line segments which can be formed using these
points.
Here’s a picture of what the circle
might look like had it included just five points. We join one point to each of the
other remaining points. We then join another of the points
to each of the remaining points and so on. Now, if we’re given a circle with
just five points on it, it’s quite straightforward to simply count the number of
line segments formed. We’re told, however, that we’re
given 87 points arranged in a circle. And so, we’re going to use the
concept of combinations here.
The combination formula gives us
the number of ways of choosing 𝑟 items from a total of 𝑛 items. It’s 𝑛 choose 𝑟, which is simply
𝑛 factorial over 𝑟 factorial times 𝑛 minus 𝑟 factorial. Now, we have a total number of 87
points, so we’re going to let 𝑛 be equal to 87. But what is 𝑟? Well, each time we draw a line
segment, we’re joining two points. So we’re going to let 𝑟 be equal
to two. We want to find the number of ways
of selecting any two points from the given 87. And so, the number of line segments
which can be formed is given by 87 choose two.
According to the formula, that’s 87
factorial over two factorial times 87 minus two factorial, which simplifies to 87
factorial over two factorial times 85 factorial. Now, we know that 87 factorial is
87 times 86 times 85 and so on. Two factorial is two times one,
which is just two. And 85 factorial is 85 times 84 and
so on. We really do want to try and avoid
evaluating these factorials though. So instead, we spot that we can
write 87 factorial as 87 times 86 times 85 factorial. Then, we divide through by 85
factorial. We also notice that we can divide
through by two. And 87 choose two simplifies to 87
times 43. 87 times 43 is 3,741. And so, there are 3,741 line
segments that can be formed using these points.
Now, interestingly, that wasn’t the
only method we could have used. Let’s go back to the circle which
contains five points on its circumference. We started by drawing four line
segments. Remember, we picked a point on the
circle and then joined it to the remaining four. Then, we picked another point and
joined it to the remaining three. We’d already joined it to the first
point, so we reduce this number by one. We pick another point. It has already been joined to two
of the points, so it only gets joined to two more. Finally, we have one point left
over which can only be connected to another point. So, given five points arranged in
the circle, the number of line segments is four plus three plus two plus one.
We can generalize this and say,
given 𝑛 points in a circle, the sum would be 𝑛 minus one plus 𝑛 minus two plus 𝑛
minus three and so on. And so, for 87 points, the sum
would be 86 plus 85 plus 84 and so on. Notice that each number in this sum
forms a sequence. In fact, it’s an arithmetic
sequence. And so, we can use the formula for
the sum of the first 𝑛 terms of an arithmetic sequence. That’s a half 𝑛 times two 𝑎 plus
𝑛 minus one 𝑑, where 𝑎 is the first term in the sequence and 𝑑 is the common
difference.
Our sequence, in fact, is a series
because we’re finding the sum of the terms has 86 terms. The common difference is negative
one. Each term reduces by one each
time. The first term 𝑎 is 86. So we get a half times 86 times two
times 86 plus 86 minus one times negative one, which also gives us 3,741.
