Combinatorics problem(board)

Mathegocart · 2017-02-02 22:55:10

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)?

bobbym · 2017-02-03 05:41:33

Hmmm, you see lots of people who try to solve combinatorics like that. And usually it works. But maybe we can do a bit better than usually.

When we solve a quadratic using the formula we are encouraged by our teachers to plug in those answers to check. This is a good rule and weeds out errors. Suddenly, when those same guys do a combinatoric problem they are satisfied with not checking. They just put down 200 or 300 binomials, a couple of factorials, get some answer, sit back and light up their pipe convinced that they are right.

EM guys are different, they want to check all answers. They want to weed out errors. They do not smoke pipes. So supposing I said to you that answer is kaboobly doo. You of course would like to prove to me it is right. How would you do that?

haidi · 2017-02-04 01:06:28

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

So, we've got two points. One is chosen out of 49, the other out of the remaining 48. With no additional constraints, that's 49*48=2352.

I suppose "adjacent" here means just touching (I'm going to continue assuming it means both side-to-side or corner-to-corner touching; you can then try doing on your own just for side-to-side). Then, the first point can be a corner-point (4 different points) or an edge-point (excluding 4 corners - 20 different points) or an inside-point (excluding edges and corners - 25 different points).

There are 3 possible scenarios:

A) If our first chosen point is a corner-point, it would have 3 adjacent points
B) If our first chosen point is an edge-point, it would have 5 adjacent points
C) If our first chosen point is an inside-point, it would have 8 adjacent points

All this you need to take into account.

In a case A scenario, the probability is 4/49*3/48;
In a case B scenario, the probability is 20/49*5/48;
In a case C scenario, the probability is 25/49*8/48.

Then, we sum the results in all 3 scenarios and we've got the final answer.

Hope it helps.

bobbym · 2017-02-04 03:11:45

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...

haidi · 2017-02-04 03:32:01

bobbym wrote:

It all depends on how he interprets the word adjacent...

I simply assumed he did not interpret it well, got the wrong answer, and asking someone to help him out now. I mean, why would he ask for help if he got it right in the first place? Doesn't make much sense to me...

bobbym · 2017-02-04 03:34:17

Hi;

Only if he calls adjacent squares ones that share sides which I believe he did.

Sometimes people just want their work checked and on other forums it is mandatory to show all your work.

haidi · 2017-02-04 04:02:47

bobbym wrote:

Only if he calls adjacent squares ones that share sides which I believe he did.

Still, if they share only sides, a corner-point has 2 adjacent points, an edge-point has 3, and an inside one has 4. Maybe, in the end, I don't understand the meaning of the word "adjacent", being a non-native English speaker...

Never mind, I didn't mean to interfere, just wanted to help him out a bit. Sorry if I caused any confusion.

bobbym · 2017-02-04 09:34:57

Hi;

No confusion at all. But what did your method get for an answer?

haidi · 2017-02-04 10:48:21

Hi,

I explained my method in post #3. If adjacent means both side-to-side contact and corner-to corner contact, I get 312/2352; if it is just side-to side, then 168/2352.

Mathegocart · 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.

Mathegocart · 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)

bobbym · 2017-02-04 11:50:55

Hi;

Okay, everybody agrees.

Math Is Fun Forum

#1 2017-02-02 22:55:10

Combinatorics problem(board)

#2 2017-02-03 05:41:33

Re: Combinatorics problem(board)

#3 2017-02-04 01:06:28

Re: Combinatorics problem(board)

#4 2017-02-04 03:11:45

Re: Combinatorics problem(board)

#5 2017-02-04 03:32:01

Re: Combinatorics problem(board)

#6 2017-02-04 03:34:17

Re: Combinatorics problem(board)

#7 2017-02-04 04:02:47

Re: Combinatorics problem(board)

#8 2017-02-04 09:34:57

Re: Combinatorics problem(board)

#9 2017-02-04 10:48:21

Re: Combinatorics problem(board)

#10 2017-02-04 11:04:51

Re: Combinatorics problem(board)

#11 2017-02-04 11:09:54

Re: Combinatorics problem(board)

#12 2017-02-04 11:50:55

Re: Combinatorics problem(board)

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