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

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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?

\(A. 150\)
\(B. 180\)
\(C. 190\)
\(D. 210\)
\(E. 240\)
[Reveal] Spoiler: OA

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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
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)

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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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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] 21 Mar 2018, 02:51
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