In how many ways can we choose two black squares on a chess board

Question

In how many ways can we choose two black squares on a standard 8x8 chess board, so that they do not lie in the same row or column?

A. 240

B. 400

C. 496

D. 788

E. 1024

CrackverbalGMAT · Accepted Answer

Method 1:

Total number of black squares on the board = 32
Number of ways of selecting the first black square =  32C_1  = 32
The selected black square will be common in the row and column from which we cannot select the second black square, 
Therefore, the number of black squares  unavailable  for selecting the second = 4 in the row which cannot be used + 4 in the column which cannot be used - 1 common = 7
Therefore, number of black squares available for selecting the second = 32-7 = 25
Number of ways of selecting the second black square =  25C_1  = 25

Therefore, total number of ways of selecting the 2 black squares = 32*25/(No. of repeating colours!) = 800/2! =  400  (we divide because the two squares are both black and hence the order of selection doesn't matter here)

Answer (B)

Method 2:

Number of ways of selecting 2 black squares from the same row =  4C_2*8  = 48
Number of ways of selecting 2 black squares from the same column =  4C_2*8  = 48
Total number of  incorrect  ways of selecting 2 black squares (ie; such that they either lie in the same row or the same column) = 48 + 48 = 96
Total number of ways of selecting 2 black squares =  32C_2  =  \frac{32*31}{2*1}  = 496

Therefore, total number of ways of selecting the 2 black squares such that they don't lie in the same row or column = 496 - 96 =  400

Answer (B)

Hope this helps.

AnuragTiwari1991 · Answer

svasan05

In method 1 there is a calculation mistake 32*25 <> 400. Thanks for method 2, a good approach

CrackverbalGMAT · Answer

Thanks Anurag, for pointing out.

I had missed out an important step. We must divide the final answer by 2 since the black squares are the same kind and hence the order of appearance doesn't matter (this is similar to forming words with repeating letters). This should lead to the correct answer.

I'll edit the solution to reflect this.

Elessar2478 · Answer

Could you please explain why you've divided by 2?

To me it seems like the answer should be = (ways of selecting first square)*(ways of selecting second square once the first one has been selected)= 32*25 = 800.

How does the fact that both squares are black, and hence the order of selection doesn't matter, post into this. I mean, I don't see the step where we counted the possibilities twice (due to order of selection) because of which we're now having to divide by 2 at the end.

Posted from my mobile device

nick1816 · Answer

As second square is not in the row or column of first square, the first square is not in the row or column of second square either. Hence, we counted any pair twice. We have to divide by 2.

MathRevolution · Answer

There are 64 squares in total with 8*8 rows and columns.

64 squares(32 Black + 32 White).

So every Row/Column will have 4 black squares. (Note: 1 square selected will always be common for both row and column and as both the squares are black in color, we need to remove the duplicate selection, and hence the total result will be halved.).

1st Black square can be selected out of any 32 squares in 32 ways.

2nd Black square: Do not lie in the same row or column?

Attachment:

chessboard.png [ 10.03 KiB | Viewed 12569 times ]

Suppose you selected 1st Black square from Row1/Column1: This implies the remaining 3 squares from the same row and 3 squares from the same column cannot be taken + the same square which is already selected. So, we cannot take 7 squares out of 32 squares, if we choose 1 square. So, the total possible squares available for the selection are 32-7 = 25(Square means Black squares). Also, selecting a row will affect the column or vice-versa.

So, total combinations will be 32*25 / 2 = 400.

Answer B

GMATinsight · Answer

In a chess board there are 64 squares

32 are white and 32 are black

Each row and column has 4 black square

First square can be chosen in 32 ways

Now the second block can NOT be chosen from the same row or columns so 7 black squares can't be used for next selection hence options left = 25

Total Possible outcomes = 32*25 / 2! =400

P.S. Why did I divide it by 2!? because the step by step selection includes arrangement whereas this question is simply about selection of two black squares. The arrangement of two blocks can be counted by 2! ways and teh effect of arrangement is excluded by dividing the result by 2!

Hope this help!

In how many ways can we choose two black squares on a chess board

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In how many ways can we choose two black squares on a chess board

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