Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
09 Jul 2025, 22:35
Kudos
Bookmarks
Probability that all choose the same number:
Each person independently picks from 1 to n -> so there are total n^n total possible outcomes.
For all to pick the same number, everyone must pick 1, or 2, ..., up to n -> n favorable cases (one for each number).
Psame = n / (n^n)
[hr]
Step 2: Probability that all choose different numbers:
The number of favorable cases = number of permutations of n distinct numbers:
Pdiff = n! /n^n
[hr]
Step 3: Use the given condition:
n! / (n^n) = 6 * n / (n^n)
Multiply both sides by n^n:
n! = 6*n
Now solve for n such that:
n!= 6n => n! / n = 6
=> (n−1)! = 6 (3! => 3*2*1 => 6 )
=> n−1 = 3
=> n=4 (B)
Each person independently picks from 1 to n -> so there are total n^n total possible outcomes.
For all to pick the same number, everyone must pick 1, or 2, ..., up to n -> n favorable cases (one for each number).
[hr]
Step 2: Probability that all choose different numbers:
The number of favorable cases = number of permutations of n distinct numbers:
Pdiff = n! /n^n
Step 3: Use the given condition:
n! / (n^n) = 6 * n / (n^n)
Multiply both sides by n^n:
n! = 6*n
Now solve for n such that:
n!= 6n => n! / n = 6
=> (n−1)! = 6 (3! => 3*2*1 => 6 )
=> n−1 = 3
=> n=4 (B)
Bunuel
09 Jul 2025, 22:51
Kudos
Bookmarks
Bunuel
Total cases to choose n numbers from 1 to n = n^n.
all of them choose different numbers = n!
Probability when all of them choose different numbers= n!/n^n.
All of them choose the same number: they can choose n numbers
all choose 1...
all choose 2.....
................
all choose n...
cases = n
Probability when all of them choose the same number = n/n^n= 1/n^(n-1).
Given that
Probability when all of them choose different numbers= 6 * Probability when all of them choose the same number
n!/n^n= 6 * /n^(n-1).
n (n-1)! /n^n = 6n/ n^n
(n-1)! = 6
3 * 2 * 1 = 6
n-1=3
n=4
09 Jul 2025, 23:08
Kudos
Bookmarks
Each of n people randomly selects an integer from 1 to n, inclusive. If the probability that all of them choose different numbers is six times the probability that all of them choose the same number, what is the value of n?
n People, 1 to n numbers.
P(ndiff) = 6*P(nsame)
P(ndiff) = n! / n^n (since each person can choose one number and then that no is removed from the pool and the next person can choose only from n-1 numbers and so on)
P(nsame) = n / n^n (since we just need to choose which is the one number every person is going to select from the n given numbers, that is just n ways)
Now, putting together:
n! / n^n = 6*(n / n^n)
=> n! / n = 6
=> (n-1)! = 6 = 3!
=> n-1 = 3
=> n = 4
Answer is B
