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PROBLEM WITH CHESS BOARD

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PROBLEM WITH CHESS BOARD

by candrapetra » Mon Aug 22, 2011 4:20 pm
I'm working on a word problem that I can't seem to wrap my head around. Can someone please breakdown how to solve this? Thanks!

How many ways are there to choose two different squares on a chess board so that they are not in same colomn?

If, in addition, they are not on the same diagonal?

What about on an n X n chess board ?
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by sumgb » Tue Aug 23, 2011 2:12 pm
For 8 x 8 chess board,

you can select two columns in 2* 8c2 ways (since order doesnt matter).
in these 2 columns, you can have 8 * 7 squares selection so IMO the answer should be

2 * 8c2 * 8 * 7 = 112 * (8! / (6!*2!)) = 112 * 28 = 3136 ways

can you please post the answer here?
hope this helps..
Thanks,
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by akhilsuhag » Tue Aug 23, 2011 2:28 pm
How many ways are there to choose two different squares on a chess board so that they are not in same colomn?

I think this is how to do it for an 8*8:

Select one Column AND Select 2 squares on that column

=> Select a column = 1C8 (we have 8 columns and need to select 1) = 8
=> Select 2 sq on a column = 2C8 = 28

So, total number of ways = 28 * 8 = 224

Plz can you post the OA.

If, in addition, they are not on the same diagonal?

I don't understand this part of the question, if they are on the same row/column then they can never be on the samr diagonal.

What are the origins of the question. I mean whr did you find it??
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by Frankenstein » Tue Aug 23, 2011 7:24 pm
candrapetra wrote: How many ways are there to choose two different squares on a chess board so that they are not in same column?
Hi,
Out of the 64 squares, 1st square can be picked in 64C1 = 64 ways.
2nd square should be picked from different column. So, it should be picked from 56 squares in 56C1 = 56 ways.
As every pair is counted twice, we divide the total number by 2.
So, total number of ways = 64*56/2 = 1792.

@akhilshuag I guess you misread the question. It is asking about squares not in same column. You have calculated for squares in same column. So, all you need to do is subtract that from 64C2.
So, answer will be 64C2 - 8*8C2 = 1792.
Cheers!

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by Frankenstein » Tue Aug 23, 2011 7:29 pm
sumgb wrote:For 8 x 8 chess board,

you can select two columns in 2* 8c2 ways (since order doesnt matter).
in these 2 columns, you can have 8 * 7 squares selection so IMO the answer should be

2 * 8c2 * 8 * 7 = 112 * (8! / (6!*2!)) = 112 * 28 = 3136 ways

can you please post the answer here?
hope this helps..
Thanks,
Hi,
Since, order doesn't matter, it should be 8C2. You don't need to multiply by 2. If you are multiplying by 2, you are essentially taking order into consideration.
After the columns are selected, you have 8*8(not 8*7, because they can be in same row but in different columns) squares for selection.
So, total number of ways is 8C2 *8*8 = 1792.
Cheers!

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by akhilsuhag » Wed Aug 24, 2011 12:49 am
@akhilshuag I guess you misread the question. It is asking about squares not in same column. You have calculated for squares in same column. So, all you need to do is subtract that from 64C2.
Yeah I sure did. Really stupid of me. It is a perennial problem that I am dealing with, I assume the question and get it wrong.

Thanks for pointing out!!
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by quantskillsgmat » Wed Aug 24, 2011 1:58 am
64^2+63^2.......1^2 is answer.
for any nxn boxes number of squares is summation of n^2 where n is 1 to n
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by saketk » Wed Aug 24, 2011 9:31 am
quantskillsgmat wrote:64^2+63^2.......1^2 is answer.
for any nxn boxes number of squares is summation of n^2 where n is 1 to n
Wrong answer! This equation will give you a very very biiiig answer. For this question you just have to use selection method
A chess board consist of 64 squares. First block can be selected in 64 ways
Now, as per the condition given - we cannot select a block from the same column. Hence the number of blocks left are 56 only

Find the product and divide by 2. Why 2? Because the pair AB is similar to BA. Hence we have to remove the repetitive count. This will give you the correct answer
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by thestartupguy » Sat Sep 03, 2011 3:33 pm
I agree with Frank to a certain extent. The foll. is the correction to his answer

Hi,
Out of the 64 squares, 1st square can be picked in 64C1 = 64 ways.
2nd square should be picked from different column. So, it should be picked from 49 (not 56) squares in 49C1 = 49 ways.
As every pair is counted twice, we divide the total number by 2.
So, total number of ways = 64*49/2 = x (can be solved)

Generalizing, for n x n squares, n^2 x (n-1)^2/2

Not on the same diagonal.

IMO the answer is 64 x 56 /2 or n^2 (n^2-n)/2
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