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Bunuel
How many pairs of integers (x, y) with x > y can be selected from the range -5 to 4, inclusive?
A. 15
B. 18
C. 36
D. 45
E. 90

Sol: IMO-Option-D

Between -5 to 4, total no.of Integers = 4-(-5)+1 = 10
Any pair selected from the 10 nos. will have one of the numbers greater than the other.
To count it once (for x > y) , take combination (rather than all arrangements)

Total no. of ways to select 2 nos. out of 10 = 10C2 = 10!/(10-2)!2! = 45
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An Simple solution would be:

If we choose any two different numbers one will be greater another will be less .

So X > Y and Y > X are equally probable and distributed

There are total of 100 pairs of numbers (X , Y) i.e 10 * 10

Of those 10 pairs i.e (-5, -5) , (-4, -4) ... ( 4,4 ) are equal pairs
Out of 90 that are left they will be equally probable to be either X > Y or x < Y.


Hence 45 satisfy X > Y condition.
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