Re: number of all possible ordered pairs
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18 Feb 2022, 16:58
Approach -
Step 1 : Find total number of terms in Set E -
Identified that all numbers have a difference of "4"
If we subtract the set by "1" we get,
(1-1) = 0 = 4*0
(5-1) = 4 = 4*1
(9-1) = 8 = 4*2
.
.
.
(497-1) = 496 = 4*124
Thus, number of terms = 124 + 1 (term for "0" position) = 125
Step 2 : Identify how many pairs can be formed
We can see that when we try to create pairs where first number is greater than the second the total number of combinations for n numbers is always {summation of first (n-1) numbers}
If this does not strike out to you, just use a simple case of taking 5 numbers : 1, 2, 3, 4, 5
Where we see pairs are -
(5,1), (5,2), (5,3), (5,4)
(4,1), (4,2), (4,3)
(3,1), (3,2)
(2,1)
Thus number of pairs are {summation of (5-1)} = 4 + 3 + 2 + 1 = 10
Similarly here we get,
Total number of pairs = {summation of (125-1)} = {summation of (124)}
using formula of {summation(n)} = n(n+1)/2
We get,
{summation of (124)} = 124*125/2 = 7750
Thus the answer is C