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# In how many ways can two integers m and n, with m > n, be selected fro

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5839
GMAT 1: 760 Q51 V42
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In how many ways can two integers m and n, with m > n, be selected fro [#permalink]

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19 Mar 2018, 02:15
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Difficulty:

35% (medium)

Question Stats:

73% (01:25) correct 27% (01:28) wrong based on 52 sessions

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[GMAT math practice question]

In how many ways can two integers $$m$$ and $$n$$, with $$m > n$$, be selected from the whole numbers from $$12$$ to $$32$$, 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 $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" BSchool Forum Moderator Joined: 26 Feb 2016 Posts: 2939 Location: India GPA: 3.12 In how many ways can two integers m and n, with m > n, be selected fro [#permalink] ### Show Tags 19 Mar 2018, 02:25 MathRevolution wrote: [GMAT math practice question] In how many ways can two integers $$m$$ and $$n$$, with $$m > n$$, be selected from the whole numbers from $$12$$ to $$32$$, inclusive? $$A. 150$$ $$B. 180$$ $$C. 190$$ $$D. 210$$ $$E. 240$$ There are $$21(32 - 12 + 1)$$ integers between 12 and 32. Since the integers, m and n need to be selected from the 21 numbers available, there are a total of 21*20 ways of choosing these integers. Of these integers, half the possibilities satisfy the condition m > n. Therefore, there are $$\frac{21*20}{2} = 210$$ ways of selecting these integers (Option D) _________________ You've got what it takes, but it will take everything you've got CEO Joined: 12 Sep 2015 Posts: 2633 Location: Canada Re: In how many ways can two integers m and n, with m > n, be selected fro [#permalink] ### Show Tags 19 Mar 2018, 08:15 1 Top Contributor MathRevolution wrote: [GMAT math practice question] In how many ways can two integers $$m$$ and $$n$$, with $$m > n$$, be selected from the whole numbers from $$12$$ to $$32$$, inclusive? $$A. 150$$ $$B. 180$$ $$C. 190$$ $$D. 210$$ $$E. 240$$ A nice rule says: the number of integers from x to y inclusive equals y - x + 1 32 - 12 + 1 = 21 So, there are 21 numbers from which to choose NOTE: the order in which we select the numbers does not matter because, once we have selected the 2 numbers, we'll let m equal the larger value (to maintain the condition that m > n) Since order does not matter, we can use combinations. We can select 2 numbers from 21 numbers in 21C2 ways 21C2 = (21)(20)/(2)(1) = 210 Answer: D RELATED VIDEO - calculating combinations (like 21C2) in your head _________________ Brent Hanneson – Founder of gmatprepnow.com Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5839 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: In how many ways can two integers m and n, with m > n, be selected fro [#permalink] ### Show Tags 21 Mar 2018, 02:51 => Since the order of $$m$$ and $$n$$ is fixed, we only need to count the number of ways to choose $$2$$ numbers from $$12, 13, …, 32.$$ We have $$21$$ numbers to choose from since $$32 – 12 + 1 = 21.$$ The number of selections is $$21C2 = \frac{(21*20)}{(1*2)} = 210.$$ Therefore, D is the answer. 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$99 for 3 month Online Course"
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Re: In how many ways can two integers m and n, with m > n, be selected fro   [#permalink] 21 Mar 2018, 02:51
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