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Re: number of all possible ordered pairs [#permalink]
could someone please this question ?
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Re: number of all possible ordered pairs [#permalink]
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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
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Re: number of all possible ordered pairs [#permalink]
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There are 125 numbers..
Number of pairs formed by 1 and any other is 124. Number of pairs formed by 5 and any other number greater than 5 is 123 and so on...
Thus giving us a total of
124+123+...+1=62(125)=7750
Answer C..

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Re: number of all possible ordered pairs [#permalink]
1
Number of elements in set

497-1= 496

496/4 + 1 =

125x124/2 = 7750

Answer C

Adewale Fasipe, quant instructor from Lagos Nigeria.
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Re: number of all possible ordered pairs [#permalink]
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