In the list: "Jumble the digits every which way: 123, 132, 213, 231, 312, 312." -- 312 is repeated twice at the end. One of these should be 321.
6
I love stuff like this! You know — math stuff. I can think of many ways this could be useful. Or at least make a good party trick.
5
I see it's finally correct. Dr. Loh is a professor at CMU, not as stated previously at Cornell.
3
“Jumble the digits every which way: 123, 132, 213, 231, 312, 312.”
I think the last number should be 321?
22
@Bobby Yup
9
"That’s combinatorics, and its useful for solving problems of this type."
Also useful: apostrophes.
61
@Cunegonde Misthaven Thanks. It’s fix.ed
21
These comments are great. They nudge at the interesting details, pose variations of the problem and help everyone appreciate the math. Thanks, everyone!
5
Actually, counting will get you somewhere in solving this problem: I cut and pasted this problem into a drawing program, made a bunch of copies of it, and then used a polygon tool to locate each and every triangle that is possible, classifying them along the way (for example, triangles that have no lines passing through them, triangles having two lines passing through, etc.) I got 20... I think I'll give this problem to my 2nd graders and see what methods they use to keep track of the triangles. Yes, combinatorics is a very clever method, but since there are only 20 triangles in all, it is not really that overwhelming; the trick is to have a system.
11
@Robert Berkman Actually, it's even easier than that: make a letter for each line (since there are 6 lines, A, B, C, D, E and F will work), then find unique combinations of the three letters:
ABC, ABD, ABE, ABF
ACD, ACE, ACF
ADE, ADF
AER
There are 4 + 3 + 2 + 1 combinations beginning with A, or 10 altogether.
Let's start with B:
BCD, BCE, BCF
BDE, BDF
BEF
or 3 + 2 + 1 = 6.... I wonder where this is going?
9 starting with A, 6 starting with B, hmmmm.....
CDE, CDF
CEF
2 + 1 = 3, so the pattern is 9, 6, 3.....
Then our last combination: DEF
9 + 6 + 3 + 1 = 20
I know it's fun to show off our knowledge of high level mathematics, but this is really a problem that just requires a single insight to "crack" it. One you have it, it's just a matter of making combinations, looking for patterns and then using that to get the answer.
Indeed, my second grade students will love solving this!
8
@Robert Berkman This is what I get for doing math at 1:30 am.... the sequence is 10, 6, 3, 1, not 9, 6, 3, 1.... which makes it more interesting, because 1, 3, 6 and 10 are the first 4 triangular numbers, and because the sequence is 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10.... which means if I wanted to do 7 lines, could I just add on the next triangular number, which is 10 + 5 = 15, so 1 + 3 + 6 + 10 + 15 = 35 triangles possible if there are 7 lines instead of 6?
2
You know there's a formula for this right? It is m! / (n!)(m-n)! where m is the total number of objects (in this case there are six lines) and n is the number that are being chosen at one time (in this case you need 3 at a time to make a triangle). The exclamation mark is the mathematical symbol for factorial. 6! is 6x5x4x3x2x1 or 720. 3! is 3x2x1 or 6 and (6-3)! is also 6 so that resolves to 720/ 36 which is 20. Once you have the formula, the problem is actually not very interesting.
10
I did this in a slightly different manner. Each line is intersected five times so there are four internal segments which can serve as a base for a triangle. Combining adjacent segments and not double counting gives 4 + 3 + 2 +1 = 10 possible bases per line. There are six lines, so this gives 60 possible bases total. But each base also serves as a side on two other triangles, so one has to divide by 3 to get the unique number of triangles.
2
"Each line is intersected five times ..."
More generally, you would need to prove a statement like this before proceeding:
If there are n lines satisfying the conditions given in the article, then each line is intersected n-1 times.
In mathematical terms, that is a "lemma":
"lemma: 1: an auxiliary proposition used in the demonstration of another proposition" (merriam-webster.com)
1
Great question, but very poorly written / explained. Left out a HUGE caveat - triangles intersected by other lines, thus creating different polygons, are still counted as triangles.
7
The important thing to understand is that the solution is based on the fact that no more than two lines intersect at a single point. All lines intersect.
5
Thank-you Mr. Chang for another elucidating article on go figure. Next, could you solve the dilemma of the Grand Unified Theory. There might be a prize in it for you.
2
As far as I can see, you only get 20 if you count the open-sided angles as triangles. But why would you do that?
4
@J No, it’s closed triangles.
6
There are 20 closed triangles.
8
OK, another math major here.
Every time we sit down to dinner with guests, after we've all clicked our wine glasses, I ask everybody how many clicks were there?
Sometimes they get it right.
23
17 triangles that I saw, so pretty close.
3
Beauty. Thanks.
2
Nice problem, beautifully explained!
3
More importantly is to find an experienced tax accountant that know how to calculate the following formula:
2 + 2 = 5
2
The first explanation is terrific and easy to understand until you start exclaiming !
Then ! I start to feel a little lost !
!It's not as though I don't have an advanced degree !
I've never heard of combinatorics ! Until just now ! Sorta shocking - ( ! )
Looking up "Combinatorics For Dummies" right now ! [!]
34
@Michael c Great! Math is best if you chase your curiosity.
17
"Sorta shocking - ( ! )"
Not really. Mathematical notation has a history. The use of an exclamation mark for the factorial operation dates to the 19th century. For an extended discussion of the history of the notation for the factorial operation, see:
"A History of Mathematical Notations" by Florian Cajori.
'Looking up "Combinatorics For Dummies" right now !'
The factorial operation is discussed in the chapter on probability in this "Dummies" book:
"Finite Math For Dummies" by Mary Jane Sterling.
4
Again, the person who framed the problem has done a disservice to the mathematics. He says: “... there’s no notation at all, just a diagram and a question: How many triangles are formed by THESE six intersecting lines?” But his solution gives “how many triangle CAN BE formed...” or “how many triangles are formed by [leave out “these”] six lines...” “THESE 6 lines” is not meaningful outside of THIS diagram. No wonder people tried to count the triangles in THIS diagram. The stated solution does NOT reveal “how many triangles are formed by THESE 6 lines [in this diagram]. If you want to introduce non-mathematicians to combinatorics, you can’t blame them for not getting it, if you can’t frame the question coherently.
7
@Andy Jamieson It’s true for any six lines if a) no 2 are parallel, b) they all lie in the same plane and c) there are no points where more than 2 points intersect. That’s a lot of verbiage that is illustrated succinctly in a diagram. But there’s nothing special about the configuration shown.
15
Thank for the summer pre call refresher
Always fun
1
20 is the correct answer under a very limited set of circumstances.
And the description of those circumstances, the caption under the illustration is scarcely intelligible:
"Six lines, none parallel, and no points more than two lines intersect."
This is an illustration of what turns people off about math problems.
"20 is the correct answer under a very limited set of circumstances."
Those conditions are needed to guarantee that the number of triangles is uniquely determined by the number of lines.
"This is an illustration of what turns people off about math problems."
Math is more interesting when you realize that you don't have to accept the problem as stated, except on SAT tests. :-)
Thus, what happens if parallel lines are allowed? What happens if more than two lines can intersect at a point?
When the parallel postulate in Euclidean geometry was changed, non-Euclidean geometry was discovered.
See the Wikipedia article titled "Non-Euclidean geometry":
https://en.wikipedia.org/wiki/Non-Euclidean_geometry
There are numerous books on non-Euclidean geometry. See, for example:
"Taxicab Geometry: An Adventure in Non-Euclidean Geometry" by Eugene F. Krause.
3
The first problem is a very simple problem perhaps taught in sixth or seventh grade. It literally was no problem solving it. This particular problem pertains to ninth or tenth grade. This is taught while teaching permutations and combinations in High School and definitely interesting.
1
Wait... this isn't on the same level as the original problem, 8/2(2*2)... the original is fundamental math that everyone should know from the 6th grade... what you're figuring here is *optional* math that is generally reserved for classes with special names.
And your reason for putting it up is because mathematicians thought the previous was too rudimentary... well, the original wasn't for them... obviously.
2
@J
huh?
Basic combinatorics have application in practically every field of STEM, at the undergraduate level.
I am unaware of a field that does not teach the basic combinatorics counting math as it is fundamentally related to basic statistics (see: Statistical mechanics and thermodynamics, biological statistics, financial and economic statistics, etc.), quantum mechanics, etc.
I learned, and then relearned, that equation about 13 times as an undergraduate.
9
It’s harder but it’s something everyone can understand, maybe after some struggle, and, once you do, you have math power you didn’t have before.
The other one is like a puzzle in grammar. It might be fun but in the end you don’t get much out of it.
6
@J As I tried to explain in an earlier essay, the earlier problem was just a “gotcha,” and it really did not matter if you got the “right” answer or not. It was just showing off if you remembered this arbitrary rule or interpreted it according to a different arbitrary rule. This formula is trivial too once you learn it, but it gets used all the time.
16
A variation: what if there was one pair of parallel lines among the six?
1
Any time both parallel lines are selected, the result is not a triangle. As order is not meaningful, there are four (i.e. n-2) such combinations in all.
Answer: (6×5×4)/6-4=16
4
@RR
More elegant than my solution:
Take one of the parallel lines, and count triangles that include that line, and triangles that don't.
For the ones that don't include that line, it's the same problem with 5 lines instead of 6, so it's "5-choose-3", or 10
For the ones that do include that line, they won't include the other parallel line, so it's choosing two from the remaining four lines, or 6.
Your way is better. :-)
Here's a simple visual/counting solution. 1. Bring the image up on your computer screen and tie a piece of thread horizontal (or construct a vertical line). 2. Slide the line progressively through each intersection and count the number of intersections on each leg using the smallest number (one side might have four lines crossing, the other three so that would mean that this intersection hosts three triangles. 3. Progress one at a time and only count the triangles that lie fully in the direction of movement of your thread. In this case there are 15 stops or intersections: 4+3+2+1+0+3+3+2+0+0+1+0+1+0+0=20. I wouldn't try this solution on 7 lines though.
@Doug Leen I would put it this way: Move an imagined horizontal straightedge through the image from bottom to top, pausing at each intersection encountered to count how many lines intersect both of its upward sprouting arms. 4+3+3+2+2+2+1+1+1+1=20.
2
The combinatoric solution works unless some of the lines are parallel. As an extreme case think of three lines all parallel to each other. No. of triangles in this case is 0, not C(3,3) = 1.
2
"... unless some of the lines are parallel."
The problem statement excludes parallel lines*, although there is no reason why the problem couldn't be extended to allow parallel lines too. It's just that the number of triangles would not be uniquely determined, as you showed in your example.
* From the article: "The caveat, evident in looking the diagram, is that no two lines are parallel, and there are no points where more than two lines intersect."
Next up: Same question, but for great circles on the surface of a sphere (mathematicians like to generalize). Assume that no point of intersection has more than two great circles passing through it (this is the "generic" case).
If the meaning of "triangle" in this context isn't clear, think about that first.
1
My friend tried to manually figure out how many triangles there are but he kept going around in circles. At first glance it's hard to "square" the correct answer with the diagram but it does work. Of course if Donald Trump was trying to figure it out he would just conclude that a lot of the 20 triangles are fake.
9
"... he would just conclude that a lot of the 20 triangles are fake."
As usual, someone has to inject Trump-bashing into the comments. Anyway, Trump would say that the triangle-count is still being audited.
2
To get the answer of 20 triangles, must there also be a stipulation that the 6 lines are not in the same plane?
@P Murray:
The 6 lines are clearly in the same plane. If you can see the image as being 2-dimentional (planar), then you know they indeed ARE in the same plane.
"... a stipulation that the 6 lines are not in the same plane?"
I'm not sure what you are asking, but the lines are assumed to lie in a plane.
However, the problem could be extended to three dimensions by subdividing space with n planes and then asking how many tetrahedrons are formed.
1
@PWY
If you have a triangle in a plane, intersecting that triangle by a line in that plane, creates 2 new shapes out of the original triangle. The original triangle exists only if you ignore the intersecting line.
2
A triangle has three side and three angles. There are only six figures in the drawing provided that have those criteria all other figures have more sides and angles. So is the mathematician imagining or perceiving triangles on top of triangles? I am sorry but in basic geometrical shapes in the drawing presented there are only 6 triangles.
5
@Joe Hagy
Any one line segment can be part of the side of more than one triangle simultaneously.
4
@Joe Hagy A triangle intersected by other lines is still a triangle! The six triangles you are counting are the only ones that aren’t intersected, but that’s not what the problem asks.
8
@A "a diagram and a question: How many triangles are formed by these six intersecting lines? " Look at the diagram. How many triangles are visible. Your statement is not part of the actual question. The problem with this problem is the writing not the math or geometry. I understand what was meant. But what was meant was not what was said. It was only when the solution was explained that the underling assumptions of the question are completely revealed. Good questions are unambiguous within themselves.
5
How nice to read something that has nothing to do with the problems in the world...
40
@mj The column is certainly nice to read, but the topic it discusses, combinatorics, has a great deal to do with problems in the world, including finance and engineering risk analysis.
16
If every person and machine in the world suddenly forgot combinatorics, the world would fall apart.
3
Microscopists look for visual patterns...pencil crayons solved the problem.
3
It's a sad day when you have explain basic probability and combinatorics in such excruciating detail to a readership that in all likelihood is college educated.
12
@HistoryRhymes:
Attaining a college degree alone does not confer most of human knowledge to the degree holder. There are now so many college majors that lack any science or mathematics coursework, that they leave out vast and important areas. People who "achieve" a degree in such majors are not really educated.
This is problematic for society - since science and math form the foundations of "logic", decision-making, and critical thinking. So many lack these abilities.
24
@HistoryRhymes Yet, you might consider all of the topics that can be covered in education in the world, apply combinatorics and come to the same conclusion that there is HELLA to learn and no human could possibly touch on and remember on every single educational subset of all topics. Ever. Let me know how many topics there are, apply combinatorics and that is how many things a human would have to learn to cover it all. Its not possible.
5
Same math as you use to see that playing Powerball is for entertainment only.
17
@Eric Key
Sorry to rain on your mathematical parade, but you don't need math to determine that Powerball is for entertainment only.
All you need is to remember a quote I ran across years ago (in, IF I recall correctly, the book " Innumeracy: Mathematical Illiteracy and Its Consequences" by John Allen Paulo)
"To a mathematician, the odds of winning the lottery are not improved by buying a ticket."
While not precisely true, close enough to relegate it to the category of "entertainment only". Or as someone else once put it, "A lottery ticket is nothing more than a license to dream."
14
@Chet But it is so much more satisfying to be able to do it yourself! In the age of lies in which we live, I find myself more and more quoting Ronald Reagan "Trust but verify", or as the Russians say "Доверяй, но проверяй"
2
@Chet Read any and all books by John Allen Paulos. He’s great at explaining how math is intertwined with everything in the news and everything else too.
14
I see 8 triangles.
1
Please keep these math problem stories coming. They're a fun way to exercise my brain. Thank you.
59
I enjoy reading those "essays", but we are still (very) far from Martin Gardner (of Scientific American fame)
3
Many American have trouble figuring out the combo deals at Burger King, so I wouldn't waste my time trying to explain combinatorics to them.
3
"Many American have trouble figuring out the combo deals at Burger King, ..."
Perhaps, but many people are very good at working out baseball statistics and team configurations.
"... so I wouldn't waste my time trying to explain combinatorics to them."
With nine baseball fielding positions and a team of size n, how many possible fieldings are there? (Assuming all players are qualified for all positions.)
2
@Jay Orchard
Most of those people (all minus one?) would not try to read the article, if they happened to get near this publication.
1
I count six triangles.
1
Thank you Mr. Chang for this lesson in combinatorics, which is part of every basic course in statistics and probability. But if you have to explain to your readers what n! means, then this math lecture likely is sailing right over most readers' heads.
5
Not sure why it is difficult to count. Pretty easy to work thru visually if you start outside and move in.
@David OK. Now count it for 127 non-parallel lines.
4
@David:
As the author/article states, the underlying "problem" is not to count the triangles with one's eyes, but rather to solve the general problem for any arbitrary problem of N non-parallel lines.
The example given of 2300 lines cannot be solved accurately by most humans, using only their eyes. It can be solved - exactly - by math-literate humans, and solved very quickly and very easily using a basic calculator.
2
I printed it, got my box of colored pencils and colored each triangle in a different color.
I counted the number of pencils I used.
20. I used 20 different colored pencils.
64
@New World
I want to borrow your pencils. I counted, over & over again, and I only get 10.
4
@New World
That's impossible because many of the triangles overlap so you would have ended up with triangles of mixed up colors.
11
@RJ
There are lots of overlapping triangles. It's hard to keep track of them without tracing them in different colors. Of course, that is the whole point of combinatorics - you don't have to count them or color them to get the answer. In fact, it's black and white.
9
"The same calculation applies no matter how many lines there are."
The article could have gone on to explain that a PROOF would not use numerical examples, but algebraic notation. Further, such a proof would be by INDUCTION, which is an essential proof technique in mathematics.
Here is a simpler problem than the one in the article that is proved by induction:
Show that the sum of the natural numbers from 1 to n, where n is a natural number greater than 1, is (n(n+1))/2.
The general proof strategy is to show that the formula works for n=1 and that if the formula works for some unspecified natural number a, then the formula works for a+1.
For more about proof by induction and mathematical proofs generally, see:
* "How to prove it: a structured approach" by Daniel J. Velleman.
* "How to read and do proofs: an introduction to mathematical thought processes" by Daniel Solow.
3
While this is a nice illustration of induction, there is an easier proof. First assume n is even and (mentally) list the numbers from 1 to n. Notice that adding the lowest number (1) to the highest number (n) yields n+1. Delete the lowest and highest numbers and repeat the process, adding the result from all of the previous steps. After n/2 steps, the list will be empty, and since you have added n+1 more each time, the total will be n(n+1/2.
I've forgotten who the famous person who correctly quickly found the sum for some large number of n and was asked if he used the formula to find it. No, he replied, I added the numbers in my head.
@Mark
When I heard it long ago, the story went that this famous mathematician - I believe it was Gauss - was a kid and was misbehaving in class, he was punished by his teacher by being ordered to add up all the numbers from 1 to 1,000 and Gauss did it in a few seconds.
4
@Mark Gauss proved it as a child (and you don't need to start with an even number. Just write the sequence from left to right, then underneath that from right to left. Each column adds to n + 1, there are n columns, and you have added the numbers twice, so you and Gauss get the answer.)
2
Gah! I just had a flashback from my Intro to Stats class - hee hee - thanks for the memories!
7
I like Mr. Chang's triangle problem; much better than trivia questions about arithmetic operations. And his followup column makes an important point: a "good" solution to the triangle problem is a general solution - not just counting the triangles made by 6 lines, but coming up with an easy way to compute the number of triangles for 7, 8, 9, 100, 1000 lines.
It may be quicker to focus on the immediate problem and not think at all about the bigger world, but unfortunately that seems to be way some of our politicians operate today. "Let's create stricter laws to fight crime... but we won't worry about these laws affects the rest of society." We should have been thinking about the costs of building more prisons, spending more on police and courts, rehabilitation and training, and the social costs of unequal application of the law to poor communities.
11
Here's my question: There are 6 lines, each crosses all the other lines once--so there are thirty intersection which enclose 10 discrete shapes, 6 of them triangles, 4 rhomboids. Is this a fixed relation that will result no matter how the lines are crossed? Also, how would you determine the total number of geometric shapes that are described by the intersecting lines? And, are these shapes limited to being either 3 or 4 sided?
1
@john lafleur My guess is that you can have shapes with up to 6 sides. This is not a mathematical derivation, but rather a single possible example to visualize.
Start with a perfectly symmetrical hexagon. Then alter three of the lines by a few degrees (to comply with the no parallel lines rule. Extending all six lines indefinitely, you are still left with a hexagon "in the middle." I might well be missing something important here, but that's my attempt to answer your last question.
A musician by trade, allow me to pose another puzzler - and the answer. How many ways are there to finger a one-octave major scale, strating from the same pitch on the double bass? Answer - LOTS!
1
OK- I'll take the bait. I love the math, and get the solution you describe, but one of the beauties of such calculations is that it can be demonstrated to be true. And seeing is believing. So, let me be the troglodyte who asks where the 20 triangles are (specifically, where are the ones beyond the 5 I can clearly count)? I assume that : 1) the "triangle" must be a closed space and not simply three lines that form an open ended space, 2) the lines are infinite such that none fall outside this picture, and 3) that closed spaces that have more than 3 sides don't count. Help!
1
@Hy L. You can try out the method he describes. Pick any three lines (try to ignore the other lines). Look at the triangle formed by their intersections. Keep in mind that one of the other lines cross your triangle.
1
@Hy L. You can try out the method he describes. Pick any three lines (try to ignore the other lines). Look at the triangle formed by their intersections. Keep in mind that one of the other lines cross your triangle.
@Dan File Got it! I was looking for "clean triangles", not counting those that the other lines crossed. That was the difference. I suspect solving for what I was looking for would be far more challenging!
1
Thank you for a pleasant article that makes me smile for a change!
24
I love the verbal explanation of this puzzle, it's elegant. Combinatorics must be a fascinating study. My problem with the solution is that I'm a visual as well as verbal thinker and with my eyes I can find only six triangles in the diagram.
Who to believe, the math or my lyin' eyes?
4
"... I can find only six triangles in the diagram."
The problem statement should have made it clear* that the triangles need not all be disjoint. That is, there can be triangles within triangles. Those count as separate triangles.
That suggests a variant of the problem:
How many disjoint *polygons* are there with n lines satisfying the other conditions?
* IOW, the problem statement is still *ambiguous*.
13
@PWY
A triangle is a triangle is a triangle.
1
Kelly Logan: "A triangle is a triangle is a triangle."
That's not mathematics. In mathematics, there can be no unstated assumptions.
In fact, tilings of the plane are defined so that the individual tiles are DISJOINT. See:
"Tilings and Patterns" by Branko Grunbaum and G.C. Shephard (Dover, 2016).
I approached the problem in a different way. I selected any one line and looked at the five points of intersection. I then started at one end point and traveled to the next point of intersection. My line, the line intersecting at my starting point and the line at this next point form triangle 1. At the next point forms triangle 2 and so on. For this first line, and seemingly for all six lines, four triangles are formed making a total of 24 which was my guess. After reading the article, it appears that I have four redundancies. Now I get to figure out where they are, or else figure out the fallacy in my approach. Ain't math fun!
7
@JPZiller " Ain't math fun!" Is that a question sans proper punctuation or a statement based on opinion rather than fact?
My personal opinion is that no, math is not nor has it ever been fun. But don't take my word for it. Ask any 8 year old who ever feared his or her parents rath for failing their math grades.
1
@Martino Statement of opinion. Over the summer I had an hour with a 12 year-old who admitted she hated math. I asked her if she wanted to know how computers store data and her interest was piqued. After 90 minutes I had her fairly fluent in base 2, 8 and 16 and using that she could explain back to me why an IPv4 address can have no segment greater than 255. And believe me, that will stick because something she uses every day became fascinating via insights as to what's underneath. Or take 1,729 the taxicab number. 1+7+2+9 = 19 Reverse the digits and multiply 19 * 91 = 1,729 Math is only scary if one makes it so.
5
@Martino
Sadly, too many people think of math as simply a school subject. How pathetic would our lives be of we applied that attitude to everything we learned in school: literature, art, music, ...
1
I understand the need to use plain language but calling the permutation (6x4x3) a combination is confusing. Especially when the desired answer is a combination C(6,3).
I think what this example and the previous one with 8/2(2 2) illustrate us the need to be consistent, which math does. Math uses precise language and has well defined rules and notations. The reason is for reproducibility.
2
@NKM I actually appreciated the plain language. I'm a words-oriented person, so putting it this way made utter sense to me. Then the mathematical expression appeared, and that was far more challenging to grasp. We all approach math in different ways!
2
Key point - the lines need to be stipulated as either infinite length, or of sufficient length and placement that all intersect.
9
@Brian a line has infinite length by definition. A fine-length line is called a segment.
21
@M. While your definition of 'line' is correct, Brian's point is well taken. Just look at the picture of the lines presented by Po-Shen Loh. They are not infinite and, because of this, shifting them in the picture can result in a different number of triangles.
1
@Spatha Spatula | It is impossible to represent an infinite line graphically, not only because mortals who have finite lives cannot draw something that is infinite, but also because a graphic image doesn't correspond to what a line in mathematics is: "a straight or curved continuous extent of length without breadth."
1
Man, what does this hugely important passage mean?: “A triangle is made of three lines. So if I have six lines, then the whole question becomes how can I pick three lines out of those six lines?” What does picking three lines mean in this context?
The intrinsic difficulty of math may have to do with the complexity of putting into words concepts that are spatial and abstract. Only a tiny percentage of people have the patience and the motivation to tolerate mathematics. Tiniest still go on to become mathematicians. God bless them!
11
"What does picking three lines mean in this context?"
"Picking" is often used in mathematical proofs.
In general, "picking" means choosing an arbitrary subset of some set having specified properties.
The essential word is "arbitrary", because the subset cannot be assumed to have any special properties beyond those of the containing set.
That method is essential when proving properties of infinite sets.
For example, show that the set of natural numbers contains an infinite number of even numbers. To start the proof, *pick* an arbitrary even number and show that there is a bigger even number:
1. Pick the even number n.
2. Then n+2 is also an even number.
Filling in the details is left as an exercise. :-)
For more, see the "choose method" in
"How to read and do proofs: an introduction to mathematical thought processes" by Daniel Solow.
5
@UC Graduate
Or, for a language based explanation...
"I have 6 cards in my hand facing me. You can't tell the difference between them. Pick 3."
Now imagine all the distinct combinations of picking 3 cards out of a hand of 6. Ace, King, Jack is the same as King, Ace, Jack because you can just reorder the cards. Thats the equivalent of the triangle problem. It's noted C(6,3) or 6C3, and solves to 20.
3
Thank you for the math lesson Mr. Chang. Unfortunately, the only triangle problem that the average American is interested in solving is the Bermuda Triangle.
32
I was a math major in college and combinatorics was my favorite course. I loved its power to solve problems and I use combinatorial methods regularly to this day.
21
@Frank. I regularly favor conspiratorial methods.
5
Oh no -- here we go with the parenthesis again!
How do you separate n!/(n-r)! from the expression n!/((n-r)!r!)?
Doesn't using the double parenthesis in the denominator mean that you first perform ((n-r)! r!), and then divide n! by the result of performing the parenthetical multiplication?
1
@JM
Sure you can do that, too. But why not use the commutative property of real-number multiplication?
2
@JM You’ve got it right.
The comment to do n!/(n-r)! separately is to show you where this expression comes from. That part corresponds to picking the ordered lines. Then dividing by n! corresponds to removing the multiple orderings for the same set of lines.
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"Oh no -- here we go with the parenthesis again!"
The general expression is unambiguous. However, Ken has glossed over the implied properties of the factorial operator:
1. The factorial operator takes ONE operand -- it is a unary operator
2. The factorial operator comes AFTER the operand.
3. The factorial operator has HIGHER precedence than multiplication and division, so a/b! means a/(b!).*
Thus, the factorial operator is like the negation operator (the minus sign) in elementary arithmetic in that it is a unary operator.
However, the factorial operator is UNLIKE the negation operator, because the minus sign always comes BEFORE the operand.
Further, the notation "C(n, r)" is an example of FUNCTIONAL notation, which has its own conventions.
* See, for example, "Master math: probability" by Catherine A. Gorini, page 50.
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