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

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by Max@Math Revolution » Mon Mar 19, 2018 1:41 am
[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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by Brent@GMATPrepNow » Mon Mar 19, 2018 7:11 am
Max@Math Revolution 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

If anyone is interested, we have a free video on calculating combinations (like 21C2) in your head: [url] https://www.gmatprepnow.com/module/gmat- ... /video/789

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by swerve » Mon Mar 19, 2018 9:41 am
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 (21*20)/2 = 210 ways of selecting these integers. Option D.

Regards!
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by Max@Math Revolution » Wed Mar 21, 2018 1:44 am
=>

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

Therefore, D is the answer.

Answer: D
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by deloitte247 » Fri Mar 23, 2018 11:33 pm
The number of integer from 'x' to 'y' inclusive
=y-x+1
=32-12+1
=21
There are 21 numbers from which to choose 'm and 'n' integer and 'm' must equal to the larger value to maintain the condition that (m>n).
The order is not important combination in which 2 numbers are selected from 21 numbers is employed.
which is $$21C_2$$
$$21C_2=\frac{\left[\left(21\right)\cdot\left(20\right)\right]}{2\cdot1}=\frac{420}{2}=210\ ways\ \left(option\ D\right)$$
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