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

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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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New post 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?

\(A. 150\)
\(B. 180\)
\(C. 190\)
\(D. 210\)
\(E. 240\)

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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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New post 19 Mar 2018, 02:25
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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 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)
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Re: In how many ways can two integers m and n, with m > n, be selected fro  [#permalink]

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New post 19 Mar 2018, 08:15
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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

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Re: In how many ways can two integers m and n, with m > n, be selected fro  [#permalink]

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New post 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
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Re: In how many ways can two integers m and n, with m > n, be selected fro  [#permalink]

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New post 22 Aug 2018, 00:10
Why not answer 190? there will be 20 combinations where m=n example {12,12}, {13,13}. We need to take out those 20 combinations from 210.
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Re: In how many ways can two integers m and n, with m > n, be selected fro  [#permalink]

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New post 24 Aug 2018, 06:09
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PLUTO wrote:
Why not answer 190? there will be 20 combinations where m=n example {12,12}, {13,13}. We need to take out those 20 combinations from 210.



Hey PLUTO

Welcome to GMATClub!

We only choose different integers, as the number of ways of choosing the integers is \(\frac{21*20}{2} = 210\).

If what you said was possible, the number of ways of choosing the integers are 21*21 = 441
and we will eliminate 21 options - (12,12),(13,13).......(31,31),(32,32). This will bring down
the number of options to 441 - 21 = 420. Now, since (12,21) is the same as selecting (21,12)
we will have to eliminate those options(which are exactly half of the options). Therefore, the
total number of ways of choosing the integers are \(\frac{420}{2} = 210\)

Hope this clears your confusion.
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Re: In how many ways can two integers m and n, with m > n, be selected fro &nbs [#permalink] 24 Aug 2018, 06:09
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