Last visit was: 29 Apr 2026, 11:36 It is currently 29 Apr 2026, 11:36
Home
Messages
Tests
Schools
Events
Close
GMAT 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 GMAT 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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
User avatar
Vibhav10
User avatar
Joined: 07 Feb 2018
Last visit: 18 Feb 2022
Posts: 16
Own Kudos:
9
Given Kudos: 39
GMAT 1: 700 Q49 V37
GMAT 2: 710 Q49 V39
GPA: 4
Products:
My Reviews
GMAT 2: 710 Q49 V39
Posts: 16
Kudos: 9
Set A consists of 25 distinct numbers. We pick n numbers 26 Aug 2020, 18:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
janani
I think B is the right answer

this is my reasoning for the actual probability

Given 5 nos (doesn't matter what they are)
for the first choice- there is exactly one option out of 5 options. i.e we should pick the smallest of the nos. so probability is 1/5
for the second choice - again there is exactly one option i.e smallest no of the remanining for nos. so probability is 1/4
for the 3rd choice- only one smallest no of the remaining 3 nos. so probability is 1/3

and so

so total probability is 1/5*1/4*1/3*1/2*1 = 1/120

+1.

We should understand following two things:
1. The probability of picking any n numbers from the set of 25 distinct numbers is the same. For example if we have set of numbers from 1 to 25 inclusive, then the probability we pick n=5 numbers {3,5,1,23,25} is the same as that of we pick n=5 numbers {9,10,4,6,18}. So picking any 5 numbers \(\{x_1,x_2,x_3,x_4,x_5\}\) from the set is the same.

2. Now, imagine we have chosen the set \(\{x_1,x_2,x_3,x_4,x_5\}\), where \(x_1<x_2<x_3<x_4<x_5\). We can pick this set of numbers in \(5!=120\) # of ways and only one of which, namely \(\{x_1,x_2,x_3,x_4,x_5\}\) is in ascending order. So 1 out of 120. \(P=\frac{1}{n!}=\frac{1}{5!}=\frac{1}{120}\).

According to the above the only thing we need to know is the size of the set (n) we are choosing from the initial set A.

Answer: B.
Show more

i was a little confused can you check if my understanding is correct
no of combinations if we select n out of 25 numbers = 25Cn
these combinations can be arranged in = 25cn * n!
now one out of all the permutations for EACH combination will be in ascending order = 25cn * 1

probability = 25Cn*1/25Cn * n! = 1/n!

b gives n=5 hence b
User avatar
CEdward
User avatar
Joined: 11 Aug 2020
Last visit: 14 Apr 2022
Posts: 1,161
Own Kudos:
289
Given Kudos: 332
Posts: 1,161
Kudos: 289
Set A consists of 25 distinct numbers. We pick n numbers 15 Feb 2021, 20:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I can see why this is correct: P = 1/n! = 1/5! = 1/120, but can we also think about this in terms of binary yes/no?

The probability of picking that first number is 1.

For the second number the probability of getting a number that is part of the ascending order is 50/50 (yes/no)
third number...50/50
fourth number...50/50
fifth...50/50
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,996
Own Kudos:
1,120
Posts: 38,996
Kudos: 1,120
Set A consists of 25 distinct numbers. We pick n numbers 02 Apr 2025, 23:15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
   1   2 
Moderators:
Bunuel
Math Expert
109974 posts
kingbucky
498 posts
Gmat860sanskar
212 posts