Last visit was: 18 Dec 2024, 03:04 It is currently 18 Dec 2024, 03:04

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
ScholarDen Representative
Joined: 21 Aug 2021
Posts: 119
Own Kudos [?]: 201 [5]
Given Kudos: 6
Send PM
Intern
Intern
Joined: 02 Dec 2021
Posts: 7
Own Kudos [?]: 8 [0]
Given Kudos: 2
Send PM
ScholarDen Representative
Joined: 21 Aug 2021
Posts: 119
Own Kudos [?]: 201 [1]
Given Kudos: 6
Send PM
Intern
Intern
Joined: 01 Oct 2021
Posts: 48
Own Kudos [?]: 13 [0]
Given Kudos: 44
Send PM
Re: number of all possible ordered pairs [#permalink]
could someone please this question ?
avatar
Intern
Intern
Joined: 29 Dec 2021
Posts: 2
Own Kudos [?]: 1 [1]
Given Kudos: 1
Send PM
Re: number of all possible ordered pairs [#permalink]
1
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
Intern
Intern
Joined: 24 Mar 2020
Posts: 6
Own Kudos [?]: 8 [2]
Given Kudos: 7
Send PM
Re: number of all possible ordered pairs [#permalink]
2
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..

Posted from my mobile device
GRE Instructor
Joined: 06 Nov 2023
Posts: 88
Own Kudos [?]: 93 [1]
Given Kudos: 21
Send PM
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.
Prep Club for GRE Bot
Re: number of all possible ordered pairs [#permalink]
Moderators:
GRE Instructor
88 posts
GRE Forum Moderator
37 posts
Moderator
1115 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne