Pebble Problem

CurlyBracket · 2022-02-22 22:34:42

If 2 pebbles are placed in a random square of a square grid having 64 squares, then what is the probability of the pebbles being on the same row or column?

I have no idea as to how I should work this out. Help please!

Jai Ganesh · 2022-02-22 23:26:59

Hi

Hope it helps! (I am a bit busy now!)

zetafunc · 2022-02-25 04:23:48

Hi CurlyBracket,

First, you're being asked to consider placing 2 pebbles in an 8 x 8 grid with 64 squares. So the total number of possibilities we need to consider is '64 choose 2' - which just answers the question "how many different ways are there of choosing 2 squares out of 64"? This gives us:

Now we need to work out how many different ways two pebbles can be arranged in the same column. Well, how is this choice made? First you (a) choose one of the 8 columns that you're going to put your pebble in, and then (b) choose a square in the column you've selected to put the second pebble. There are therefore

ways to do this (you first pick one of the 8 columns, and then once you've put the pebble down you've got 7 squares left to choose from in that column).

Next, how many different ways can two pebbles be arranged in the same row? This is again 56 (just rotate the board 90 degrees and you're back to dealing with columns again).

So the probability is:

Let me know if that makes sense - happy to clarify anything if needed.

An interesting extension of this problem to check your understanding is to consider the case where you've got an n x n chessboard. What would the probability be in that case? How about the case where you have more than two pebbles? What happens if the pebbles are allowed to share a diagonal?

CurlyBracket · 2022-02-26 23:38:44

Hi zetafunc!

Thanks for the reply. I'm still new to permutation - combinations, so this seems to be a mouthful for me!

zetafunc wrote:

Is this a general formula? For example, if there are 50 cards and I have to pick any 5, then will I say:

is the no. of combinations?

I do have a few more questions, but perhaps it's best to clear them one at a time.

zetafunc · 2022-02-27 00:22:55

Hi CurlyBracket,

Yes, that's right. In general:

These are sometimes also called the binomial coefficients, because they represent the coefficients (i.e. the 'leading terms') in the expansion of A shorthand way of denoting the above is:

which you might see on your scientific calculator.

Note that an important point is whether or not the order matters. For example, suppose you have four cards labelled (A,B,C,D) and you want to choose 2 out of those 4. How many different ways are there to do it? If we don't care about the order, the combinations are just:

(AB, AC, AD, BC, BD, CD)

which represents different ways. On the other hand, if the order does matter then we'd have:

(AB, BA, AC, CA, AD, DA, BC, CB, BD, DB, CD, DC)

which represents different permutations. If the order of the objects we've selected is of interest then we use a slightly different formula:

CurlyBracket · 2022-02-27 19:31:15

Okay. So that means, if different orders are counted, we use:

And if they don't count, we use:

I think I've got it, at least till here now.

zetafunc wrote:

(a) choose one of the 8 columns that you're going to put your pebble in, and then (b) choose a square in the column you've selected to put the second pebble.

This is understandable. But where did 8×7=56 come from?

I suppose 8 stands for the number of columns and 7 stands for the number of squares left in the column after one square is filled by the first pebble. Is this right?

If so, why do multiply it?

Last edited by CurlyBracket (2022-02-28 04:46:51)

Math Is Fun Forum

#1 2022-02-22 22:34:42

Pebble Problem

#2 2022-02-22 23:26:59

Re: Pebble Problem

#3 2022-02-25 04:23:48

Re: Pebble Problem

#4 2022-02-26 23:38:44

Re: Pebble Problem

#5 2022-02-27 00:22:55

Re: Pebble Problem

#6 2022-02-27 19:31:15

Re: Pebble Problem

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