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#201 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.

#202 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.

#203 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)

#204 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.

#205 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)?

#207 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?

#209 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?

#210 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).

#212 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.

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

Stars and Bars would be applicable too, right?

#215 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.

#216 Help Me ! » Combinatorics Formula for x+y+z? » 2017-01-20 03:36:30

Mathegocart
Replies: 5

I was just contemplating this- is there a general formula for approaching problems such as "find all ordered pairs to the equation x+ y+z=q, where x,y,z and q are positive integers.".

#218 Re: Help Me ! » Geometry: Consecutive sides that are congruent not a paralellogram? » 2017-01-18 23:11:19

Never type questions at night, my question was "If two pairs of consecutive sides are congruent, then it is a parallelogram.". I wanted a specific counterexample, let me check my tinkering around for some..

#219 Help Me ! » Geometry: Consecutive sides that are congruent not a paralellogram? » 2017-01-18 15:44:21

Mathegocart
Replies: 6

If a quadrilateral has two consecutive sides that are congruent, shouldn't it be a parallelogram? Tinkering around with EM implies that it might be true, so could anyone give me a counterexample?

#220 Help Me ! » A peculiar enigma about a quadrilateral » 2017-01-18 12:52:54

Mathegocart
Replies: 1

Is it possible to create a quadrilateral with m<E=90, m<F=90, m<G= 67, and m<H = 113? Currently, a bit of tinkering around tells me no, but I would like confirmation.

#221 Help Me ! » The Perplexing Dimensions of The Rhombus. » 2017-01-16 00:57:42

Mathegocart
Replies: 2

I have been trying to solve and develop a comprehensive solution, but I haven't. I would profoundly appreciate the help from the jovial people of MIF.
HCEG is a rhombus formed by connecting the midpoints of the rectangle ABDF. Note that O is the center of the circle circumscribing ABDF, and that it is the intersection of the diagonals of the rhombus HCEG. Some dimensions: OG = 10, CK = 8.
Find a length of the side of the rhombus HCEG.
RpSMEnF.png
My diagram is currently quite convoluted and looks more unsystematic than planned out.

I did some EM work( was intuitive) and found the solution, though can't find a geometric explanation


concluded after a quick peruse of the problem/

#222 Re: Help Me ! » Find all the n numbers » 2017-01-11 22:44:54

Here's a crucial hint to prove there is only one solution: note the essential fact that if q%7D%20%5Cin%20Q

#223 Re: Help Me ! » Sandwich theorem » 2017-01-10 08:54:47

It is utilized such that limit f(x) is between two numbers.. and multiplied by another function to obtain another one.. we see the limit and see that f(x) is compressed into a number, which must be the limit
Graphically it looks quite elegant..
%28x%5E2%29sin%28x%5E%28-1%29%29.png

#224 Re: Help Me ! » Mandy Jane's Corner » 2017-01-08 14:02:57

Hi, I could help with you[Mjane] with comprehending the material, though I would like BB to specify what to teach you.

- Thanks, M.

#225 Re: Help Me ! » Logical question s » 2017-01-07 05:30:11

Zeeshan 01 wrote:

HOW TO BolD TEXT

You should probably provide your comments in a way that is suitable for conveying information.
The markup is [ b ] "something" [ / b].
(without spaces)

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