# Problem with parallel lines

1. Sep 24, 2009

### Noiro

1. The problem statement, all variables and given/known data

There are 3 parallel lines that are crossed by other 4 parallel lines. So the problem is:
Write a formula by witch you can obtain the number of all parallelograms, if m parallel lines cross n other parallel lines.

2. Relevant equations

3. The attempt at a solution

I have no idea how to find that formula... I also posted a picture so you'll know how these lines look like.
Thank you.

http://img401.imageshack.us/img401/6647/35670712.png [Broken]

Last edited by a moderator: May 4, 2017
2. Sep 24, 2009

### tiny-tim

Hi Noiro!

Try it with 3 and 4 first, to see if that gives you any ideas …

how can you count the number of parallelograms for 3 and 4, making sure that you don't miss any out?

3. Sep 24, 2009

### symbolipoint

If you were interested in the number of intersections, you would use for example, letting q = the number of intersections, q=m*n

For the parallelograms, you are loosing a count of 1 for each set of lines....
... How will this affect your formula?

4. Sep 24, 2009

### LCKurtz

Hint: How many ways can you choose the 2 horizontal lines and how many ways can you choose the 2 slanted lines to build the parallelogram?

5. Sep 25, 2009

### Noiro

I guess I'll have to subtract 1 from something in the formula, right?

LCKurtz, I think I can choose the 2 horizontal lines in 3 ways and the 2 slanted lines in 6 ways to build a parallelogram

6. Sep 25, 2009

### LCKurtz

"LCKurtz, I think I can choose the 2 horizontal lines in 3 ways and the 2 slanted lines in 6 ways to build a parallelogram"

Which gives you how many parallelograms?

7. Sep 25, 2009

### LCKurtz

This is a bit off topic, but I am wondering if someone can tell me how you get that nice graphic to show directly in your post like that without having to click on a thumbnail or a link.

8. Sep 25, 2009

### symbolipoint

That seems to be done through a PNG image stored on imageshack. Maybe uploading a picture to directly post/paste into a message gives a different result than linking to an online webhosted image?

Noiro, look carefully at the facts in your picture. You have 3 horizontal lines and 4 vertical lines; you also can count directly the 6 paralleloprams. Look how those are arranged! You can see that multiplication is embedded in the set of facts.

Further steps are that you can see 1 less row and 1 less column are the parallelogram count. Now, Do it!

9. Sep 25, 2009

### Bohrok

Click on the Insert Image button https://www.physicsforums.com/Nexus/editor/insertimage.png [Broken] when posting and paste the url of the image, or put tags around the image url.

Last edited by a moderator: May 4, 2017
10. Sep 27, 2009

### Noiro

Hmm... I just counted them don't know any other way to find the amount of them

11. Sep 27, 2009

### LCKurtz

"I just counted them don't know any other way to find the amount of them"

When you say you counted "them" do you mean the 3 and 6 answers or the number of parallelograms?

What you have so far is that each parallelogram requires two horizontal lines and two slanted lines. You have observed that you can choose the two horizontal lines in 3 ways and the two slanted lines in 6 ways. I have a couple more questions for you:

1. How did you get the answers 3 and 6? Did you use a combinations formula?
2. Given the correct answers of 3 and 6, do you see how to calculate how many parallelograms you can build?

12. Sep 27, 2009

### LCKurtz

Last edited by a moderator: May 4, 2017
13. Sep 28, 2009

### Noiro

1. No I just counted the parallelograms because I don't know the formula
2. I can find how many I can build by multiplying 3*6 so the answer is 18. But still I can't figure out that formula

14. Sep 28, 2009

### LCKurtz

$$n! = n(n-1)(n-2)...(2)(1)$$

The number of ways you can choose r things from a set on n things is sometimes called a "combination of r things from n". Two notations for it and its formula are:

$$C(n,r) = \binom{n}{r} = \frac {n!}{r!(n-r)!}$$

The 3 and 6 come from C(3,2) and C(4,2).

How would you work the problem if you had 5 horizontal and 8 slanted lines? Manually counting them isn't a good technique.

15. Sep 28, 2009

### symbolipoint

This really is easier than to deal with permutations or combinations. Back to your original example, you had 3 horizontal lines, and 4 slanted lines; let's say m=3, and n=4. How many parallelograms? Just count them, so easy to do.

Now what if you had a slightly different figure, like 2 horizontal lines and 4 slanted lines; how many parallelograms?

Let's continue this. If you have just 1 horizontal line and still 4 slanted lines, how many parallelograms?

Can you now understand how to use m and n to develop a formula for number of parallelograms?

Last edited: Sep 28, 2009
16. Sep 28, 2009

### LCKurtz

"This really is easier than to deal with permutations or combinations. Back to your original example, you had 3 horizontal lines, and 4 slanted lines; let's say m=3, and n=4. How many parallelograms? Just count them, so easy to do."

You have to be kidding. Counting them is very error prone even for 3 and 4, let alone for larger numbers. I would venture a guess most people would not find all 18 even in that small case.

17. Sep 29, 2009

### symbolipoint

You are focusing on all possible parallelograms. Maybe a formula could be found through an inductive process. The use of permutations and combinations goes beyond just the individual smallest sized parallelograms. My viewpoint was then too limited. My approach would have only yielded (m-1)*(n-1), for m and n being natural numbers larger than 1.

18. Sep 29, 2009

### tiny-tim

Try this for 3 and 4 lines …

but first let's make it easier by talking about boxes rather than lines …

the pattern is a 2x3 box …

how many 1x1 boxes are there?

how many 1x2 boxes are there?

how many 2x1 boxes are there?

how many 2x2 boxes are there?

how many 1x3 boxes are there?

how many 2x3 boxes are there?​

Then can you see any pattern that will work for a general m x n pattern?

19. Sep 29, 2009

### Noiro

LCKurtz, I am not familiar with the factorials and permutations. Can you explain me how to use them?

tiny-tim, I guess there are 6 1x1 boxes, 3 1x2 and 2x1 boxes, 1.5 2x2 boxes, 2 1x3 boxes and 1 2x3. Sorry but I can't see any pattern that will work for a general m x n pattern

20. Sep 29, 2009

### tiny-tim

Hi Noiro!
ok … why is it 6? … why is it 3? … why is it 3? (it isn't!) … why is it 1.5? (erm … how can you have 1.5 boxes? ) … why is it 2? … why is it 1?