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Preparing for the GMAT? Avoid these 5 major mistakes that can lower your score and hurt your MBA dreams. In this expert panel discussion, 4 top GMAT coaches - GMAT Ninja, Jeff Miller from TTP, Rida Shafiq from eGMAT, and Chris Gentry from Manhattan Prep- Mar 27
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19 Mar 2018, 02:15
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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?
\(A. 150\)
\(B. 180\)
\(C. 190\)
\(D. 210\)
\(E. 240\)
In how many ways can two integers \(m\) and \(n\), with \(m > n\), be selected from the whole numbers from \(12\) to \(32\), inclusive?
\(A. 150\)
\(B. 180\)
\(C. 190\)
\(D. 210\)
\(E. 240\)
19 Mar 2018, 08:15
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Bookmarks
MathRevolution
[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?
\(A. 150\)
\(B. 180\)
\(C. 190\)
\(D. 210\)
\(E. 240\)
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
General Discussion
19 Mar 2018, 02:25
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MathRevolution
[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?
\(A. 150\)
\(B. 180\)
\(C. 190\)
\(D. 210\)
\(E. 240\)
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 are selected from the 21 available numbers, there are 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)
