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#1 Re: Help Me ! » Toothpick problem: » 2017-02-16 06:11:59

thickhead wrote:

Perhaps I meant- why doesn't my strategy work?

#2 Re: Help Me ! » Toothpick problem: » 2017-02-15 22:50:18

bobbym wrote:

Hi;

I do not understand the question. How do you know how many rows there are? Or how many columns?

#3 Help Me ! » Toothpick problem: » 2017-02-15 14:52:01

Mathegocart
Replies: 8

Anyhow I solved this problem:
We connect dots with toothpicks in a grid as shown in Figure 1.14. If there are 10 horizontal toothpicks in each row and 20 vertical ones in each column, how many total toothpicks are there?
ee8309c2c96c143eddcf6169511ac63ad9979235.png
The ans says 430 toothpicks, but I my method does not concur, for some peculiar reason.
Here's how I did it: The first row of squares contains 31 toothpicks, additional ones add 21 toothpicks. Since we add 9 more, 21*9=189, so 189+31=220 toothpicks. I feel like I'm missing something crucial, but I can't figure it out currently.

#6 Re: Help Me ! » Beach volleyball - how many combinations of 5 pairs can I make from 10 » 2017-02-08 14:17:14

bobbym wrote:

Hi;

I coach beach volleyball. We field 5 pairs of athletes. How many possible combinations of 5 pairs I can make from 10 athletes?

Another GF?

#7 Re: Help Me ! » Intriguing Combinatorics problem » 2017-02-05 00:49:18

My code was:

                which generated these quintuplets..

#8 Re: Help Me ! » Intriguing Combinatorics problem » 2017-02-04 15:33:44

THat should be enthralling, let me code it tomorrow. ahhahah an A8-6410....

#9 Re: Help Me ! » Intriguing Combinatorics problem » 2017-02-04 15:08:10

What are the specifications of your desktop? I am still currently thinking of the solution(late here), did you use generating functions?

#10 Re: Help Me ! » Intriguing Combinatorics problem » 2017-02-04 12:17:57

nice? quite inelegant but gets the job done.

#11 Help Me ! » Intriguing Combinatorics problem » 2017-02-04 11:23:54

Mathegocart
Replies: 10

1. How many quintuplets of numbers (a,b,c,d,e) where a,b,c,d and e are either -1,0, or 1 such that a + b^2 + c^3 + d^4 + e^5 = 1? Any quick way to do that? had ~5 mins to do this problem and I don't think I've got all these quintuplets.

#12 Re: Help Me ! » Combinatorics problem » 2017-02-04 11:14:54

bobbym wrote:

Hi;

This will be a long reply and I still will only scratch the surface.

Depends on how you define simple. The reason for the generating function is because it eliminates the need to reason about the problem. Also, it can do much more difficult problems than this one without increasing the complexity of the solution. We turn a combinatorics problem into a computational one. Computers do not reason, they can not assist us when we reason as you did. But they can compute, which is one reason to favor the gf approach. Your computer can help you!

As for why I use them on even smaller ones is because of the teakettle principle. Why is it considered simpler to know dozens of tricks for each type of problem when the gf will do them all?

And I don't see how come it's mandatory, unless I'm missing something...

I stress it first because the gf approach is a general method for solving counting problems. The same method does them all, difficult or simple, it does not matter.

Because they can be done with a computer the answers are error free and quick. They fit nicely into a framework we call Experimental Math. Not only do we get the answer, we have proof and we know we did not make a mistake. Of course, I can use both methods and so can lots of other people, but my weapon of choice is the gf.

When I lived in Vegas I was a professional player for 20 years. I found that it was necessary on a daily basis to solve tough combinatorics and probability problems. Problems that even experienced mathematicians would sometimes get wrong. I found that a programmer who could hardly code beyond a beginners level could often solve such problems easily and quickly when math reasoning failed. It was the beginning of my education into EM.

So to sum it up, when you have to solve a quadratic which is better? Completing the square, factoring or the formula. Well, if you have been following along the answer is simple - the quadratic formula!

I have a preference for algebra instead of ugly combinatorics, gfs sound interesting to learn as a general approach.

#13 Re: Help Me ! » Combinatorics problem(board) » 2017-02-04 11:09:54

Thought process: we choose 2 blocks from 49, so 49 choose 2. Counting vertical and horizontal squares, we find 42+42=84. so 84/(49,2)

#14 Re: Help Me ! » Combinatorics problem(board) » 2017-02-04 11:04:51

bobbym wrote:

Hi;

haidi wrote:

First off, try not to think about formulas, use just plain logic. I assume bobbym is asking you to do the same.

Well not exactly, he has the right answer. It all depends on how he interprets the word adjacent... I was just trying to get him to explain a bit. Or maybe even show a bit of EM...

Adjacent isdefined as two squares bordering a side.

#15 Help Me ! » Combinatorics problem(board) » 2017-02-02 22:55:10

Mathegocart
Replies: 11

There is a 7x7 board. I choose 2 points randomly. What is the probability that     
these 2 points are adjacent?
  There are 49 choose 2 ways to choose the 2 points, and there are 42+42 possibilities for adjacent sides- so the probability is 84/(49,2)?

#17 Re: Help Me ! » Combinatorics problem » 2017-02-02 09:27:46

Yes, I had to insert another dongle... frustrating..
So I presume GFs are used for a myriad of things, including combinatorics and discrete maths in general?

#19 Help Me ! » Combinatorics problem » 2017-02-01 15:09:17

Mathegocart
Replies: 22

1. In the word "Moondust", how many 3- letter combinations are there? I said 168, since (8*7*6)/2 = 168.
The solution is 228. Could someone provide a explanation?

#20 Re: Help Me ! » The Penguin Problem: » 2017-01-29 12:44:57

1000(1.2)^(x-1), apparently, is the formula(1000 is the starting amount, and I don't know why .2 is the growth rate for these penguins.
Let me clarify by showing how the second year of penguins grow:
first: 1000, 1500 penguins, 1200 penguins(300dead)
second: 1200 penguins, 1800 penguins, 1440 penguins(360 dead)(900 males, 900 females).

#22 Help Me ! » The Penguin Problem: » 2017-01-29 04:31:19

Mathegocart
Replies: 23

Jane finds that there are 1,000 penguins on Lup Island.
Find a formula for determining how many penguins will be there in 7 years.
1) The penguin colony consists of equal amounts of male and female penguins.
2) The penguins form couples( a male and a female) and have one baby every Spring.
3) A 1/5 of the penguin population(adults and babies) dies at the end of every year.
4) 1 year old penguins will also form couples.

#24 Re: Help Me ! » Combinatorics Formula for x+y+z? » 2017-01-20 04:40:35

Stars and Bars would be applicable too, right?

#25 Re: Help Me ! » Logical question s » 2017-01-20 03:39:48

Zeeshan, in the future, it would be preferable to try LaTeX or Codecogs so people can understand your statements.

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