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24 Feb 2020, 00:22
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (02:40) correct
62%
(02:26)
wrong
based on 160
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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
A. 240
B. 400
C. 496
D. 788
E. 1024
Updated on: 22 Mar 2020, 05:01
Originally posted by CrackverbalGMAT on 24 Feb 2020, 02:43.
Last edited by CrackverbalGMAT on 22 Mar 2020, 05:01, edited 1 time in total.
Last edited by CrackverbalGMAT on 22 Mar 2020, 05:01, edited 1 time in total.
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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.
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.
General Discussion
21 Mar 2020, 23:25
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svasan05
In method 1 there is a calculation mistake 32*25 <> 400. Thanks for method 2, a good approach
In method 1 there is a calculation mistake 32*25 <> 400. Thanks for method 2, a good approach
