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

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

BrentGMATPrepNow · Accepted Answer

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
 https://www.youtube.com/watch?v=05a0A7vjG8I

pushpitkc · Answer

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)

MathRevolution · Answer

=>

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

PLUTO · Answer

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.

pushpitkc · Answer

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.

CrackverbalGMAT · Answer

Number of integers between a and b (a & b inclusive) = b-a + 1

So, number of integers b/w 12 & 32 (12 & 32 inclusive) = 32-12 + 1 = 21

We shall use combination here as the order does not matter.

For example b/w 12 and 13 the only choice for m>n is (12,13) and not (13,12) and (12,13).

So, the number of combinations possible = 21 C 2

= (21)(20)/(2)(1) = 210

(option d)
 
  D.S
GMAT SME

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

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