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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) _________________

You've got what it takes, but it will take everything you've got

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

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.
_________________

You've got what it takes, but it will take everything you've got

gmatclubot

Re: In how many ways can two integers m and n, with m > n, be selected fro &nbs
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24 Aug 2018, 05:09