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# In how many ways can two integers x and y (with x>y) be selected from

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
GMAT 1: 760 Q51 V42
GPA: 3.82
In how many ways can two integers x and y (with x>y) be selected from  [#permalink]

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03 Jan 2018, 00:43
00:00

Difficulty:

35% (medium)

Question Stats:

72% (01:46) correct 28% (01:43) wrong based on 55 sessions

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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. $$240$$

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Retired Moderator Joined: 25 Feb 2013 Posts: 1220 Location: India GPA: 3.82 In how many ways can two integers x and y (with x>y) be selected from [#permalink] ### Show Tags 03 Jan 2018, 01:55 MathRevolution wrote: [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. $$240$$ 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6815 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: In how many ways can two integers x and y (with x>y) be selected from [#permalink] ### Show Tags 04 Jan 2018, 23:57 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Joined: 30 Jan 2017
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Re: In how many ways can two integers x and y (with x>y) be selected from  [#permalink]

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05 Jan 2018, 00:13
total cases: 21c1 * 21c1 ---> 441

number of cases with x=y: 21

cases with x not equal to y: 441-21 ----> 420
As x and y have the same domain, the cases with x>y and x<y will be equal in number.
So cases with x>y are: 420/2 ---> 210

Re: In how many ways can two integers x and y (with x>y) be selected from &nbs [#permalink] 05 Jan 2018, 00:13
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