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03 Jan 2018, 01:43
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:29) correct
24%
(01:52)
wrong
based on 197
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[GMAT math practice question]
In how many ways can two integers \(x\) and \(y\) (with \(x>y\)) be selected from \(-10\) to \(10\) (inclusive)?
A. \(150\)
B. \(180\)
C. \(190\)
D. \(210\)
E. \(240\)
In how many ways can two integers \(x\) and \(y\) (with \(x>y\)) be selected from \(-10\) to \(10\) (inclusive)?
A. \(150\)
B. \(180\)
C. \(190\)
D. \(210\)
E. \(240\)
03 Jan 2018, 02:55
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MathRevolution
from \(-10\) to \(10\) there are \(21\) numbers.
if \(x\) & \(y\) are represented on a number line from \(-10\) to \(10\), then here order matters as \(y\) has to be lower than \(x\)
So total number of selection = (total number of ways of arranging \(x\) & \(y\))\(/2\) (because in half the cases \(y\) will be higher than \(x\) and we need to exclude those cases)
\(=> \frac{21_P_2}{2} = 210\)
Option D
05 Jan 2018, 00:57
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This is the number of ways of selecting two different numbers from a set of 21 numbers. The order of choosing the numbers does not matter: we simply assign the larger number to x once the choice has been made. So, the number of ways of choosing the two numbers is
21C2 = \(\frac{21*20}{(1*2)}\) = \(21 * 10\) = \(210\)
Therefore, the answer is D.
Answer: D
This is the number of ways of selecting two different numbers from a set of 21 numbers. The order of choosing the numbers does not matter: we simply assign the larger number to x once the choice has been made. So, the number of ways of choosing the two numbers is
21C2 = \(\frac{21*20}{(1*2)}\) = \(21 * 10\) = \(210\)
Therefore, the answer is D.
Answer: D
