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 ?
PROBLEM WITH CHESS BOARD
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- candrapetra
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- sumgb
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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,
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,
- akhilsuhag
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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??
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??
Please press "thanks" if you think my post has helped you.. Cheers!!
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Hi,candrapetra wrote: How many ways are there to choose two different squares on a chess board so that they are not in same column?
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!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
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Hi,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,
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!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
- akhilsuhag
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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.@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.
Thanks for pointing out!!
Please press "thanks" if you think my post has helped you.. Cheers!!
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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
for any nxn boxes number of squares is summation of n^2 where n is 1 to n
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Wrong answer! This equation will give you a very very biiiig answer. For this question you just have to use selection methodquantskillsgmat 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
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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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
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