Last visit was: 21 Apr 2026, 02:27 It is currently 21 Apr 2026, 02:27
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
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,715
Own Kudos:
810,345
 [326]
Given Kudos: 105,795
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,715
Kudos: 810,345
 [326]
If 5 integers are to be selected at random and without replacement 14 Feb 2024, 07:41
11
Kudos
Add Kudos
313
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

35% (medium)

Question Stats:

79% (02:18) correct 21% (02:34) wrong based on 1830 sessions

History

Date
Time
Result
Not Attempted Yet
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Show Spoiler
­
ShowHide Answer
Official Answer
D
Using official mocks already? Add GMAT Club Tests for harder practice, analytics, and explanations without overlapping Official Guide or official mock questions. Try it now →
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,715
Own Kudos:
810,345
 [98]
Given Kudos: 105,795
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,715
Kudos: 810,345
 [98]
If 5 integers are to be selected at random and without replacement 14 Feb 2024, 07:49
39
Kudos
Add Kudos
59
Bookmarks
Bookmark this Post
 
Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

 
Show more
Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:
  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 18/20 * 17/19 * 16/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:

\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
User avatar
GMATGuruNY
User avatar
Joined: 04 Aug 2010
Last visit: 02 Apr 2026
Posts: 1,347
Own Kudos:
3,904
 [55]
Given Kudos: 9
Schools:Dartmouth College
Expert
Expert reply
Posts: 1,347
Kudos: 3,904
 [55]
If 5 integers are to be selected at random and without replacement 23 Feb 2024, 08:10
43
Kudos
Add Kudos
12
Bookmarks
Bookmark this Post
Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60
Show more
If 5 distinct integers are selected from the first 20 positive integers, what is the probability that 10 and 20 are among the 5 integers selected?

A pool of 20 integers, 5 slots to fill.
The probability that 10 occupies one of these 5 slots = 5/20
19 integers remain, 4 slots left to fill.
The probability that 20 occupies one of these 4 slots = 4/19
Thus:
P(both 10 and 20 are selected) = 5/20 * 4/19 = 1/19

Show Spoiler
D
­
User avatar
GMATGuruNY
User avatar
Joined: 04 Aug 2010
Last visit: 02 Apr 2026
Posts: 1,347
Own Kudos:
3,904
 [26]
Given Kudos: 9
Schools:Dartmouth College
Expert
Expert reply
Posts: 1,347
Kudos: 3,904
 [26]
If 5 integers are to be selected at random and without replacement 16 Feb 2024, 10:58
16
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
 
Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60


 
Show more
\(­P = \frac{good-outcomes}{all-possible-outcomes}\)

All possible outcomes = all possible 5-integer combinations:
From 20 integers, the number of ways to choose 5 = 20C5 = \(\frac{20*19*18*17*16}{5*4*3*2*1}\)­

Good outcomes = 5-integer combinations that include 10 and 20:
We must select 3 integers from 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, and 19 to combine with 10 and 20.
From the 18 integers in blue, the number of ways to choose 3 to combine with 10 and 20 = 18C3 = \(\frac{18*17*16}{3*2*1}\)

To divide the second fraction by the first, multiply the second fraction by the reciprocal of the first:
\(\frac{18*17*16}{3*2*1} * \frac{5*4*3*2*1}{20*19*18*17*16} = \frac{5*4}{20*19} = \frac{1}{19}\)­­­­­­­

Show Spoiler
D
­­­­
General Discussion
User avatar
QuantMadeEasy
User avatar
Joined: 28 Feb 2014
Last visit: 01 Mar 2026
Posts: 502
Own Kudos:
801
 [11]
Given Kudos: 78
Location: India
Concentration: General Management, International Business
GPA: 3.97
WE:Engineering (Education)
Posts: 502
Kudos: 801
 [11]
If 5 integers are to be selected at random and without replacement Updated on: 15 Jul 2024, 13:38
Originally posted by QuantMadeEasy on 16 Feb 2024, 02:32.
Last edited by QuantMadeEasy on 15 Jul 2024, 13:38, edited 1 time in total.
6
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
5 integers are to be selected at random, as the order of selection does not matter, the question is on combination
Total number of ways in which 5 integers can be selected out of 20 integers is 20C5

Integers 10 and 20 to be there means remaining 3 integers to be selected out of 18 integers.
Total number of ways in which 3 integers can be selected out of 18 is 18C3

Required probability = 20C5/ 18C3
Simplifying to get 1/19

IMO D

For complete pocket friendly GMAT Quant video course: 
https://www.udemy.com/course/best-gmat- ... 37602183BB
­
User avatar
kevinhirose
User avatar
Joined: 03 Jul 2023
Last visit: 28 Jun 2024
Posts: 7
Own Kudos:
4
 [2]
Given Kudos: 253
Posts: 7
Kudos: 4
 [2]
If 5 integers are to be selected at random and without replacement 22 Feb 2024, 21:44
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Hi Bunuel

Could you explain to me why we can’t do 2/20 * 1/19 * 1 * 1 * 1? Isn’t it 2/20 because we can either choose 10 or 20 and then it is 1/19 because after we choose 10 or 20 it’s whichever one is left?

I do understand how to do 18C3
/ 20C5 but I also get confused sometimes on problems like this and solve it wrong.

Posted from my mobile device
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,715
Own Kudos:
810,345
 [3]
Given Kudos: 105,795
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,715
Kudos: 810,345
 [3]
If 5 integers are to be selected at random and without replacement 23 Feb 2024, 01:21
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kevinhirose
Hi Bunuel

Could you explain to me why we can’t do 2/20 * 1/19 * 1 * 1 * 1? Isn’t it 2/20 because we can either choose 10 or 20 and then it is 1/19 because after we choose 10 or 20 it’s whichever one is left?

I do understand how to do 18C3
/ 20C5 but I also get confused sometimes on problems like this and solve it wrong.

Posted from my mobile device
Show more
­The reason why using 2/20 for selecting either 10 or 20 and then 1/19 for selecting the remaining one is incorrect is because it results in double-counting certain outcomes.

For instance, selecting 10 first and then 20 is one outcome, and selecting 20 first and then 10 is another outcome. However, both outcomes essentially represent the same event: choosing both 10 and 20. By using 2/20, we're inadvertently counting this event twice.
User avatar
einstein801
User avatar
Joined: 23 Jan 2024
Last visit: 18 Feb 2025
Posts: 152
Own Kudos:
224
Given Kudos: 138
Posts: 152
Kudos: 224
If 5 integers are to be selected at random and without replacement 14 Mar 2024, 02:04
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel, what is wrong with 1 - 18C5/20C5?

I take 18C5 to be choose 5 numbers when 10 and 20 are excluded. So in the equation 1 - 18C5/20C5 --> the solution gets us the probability when 10, 20 are included.

Thank you in advance for your kind response.
Bunuel

Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Show Spoiler
­
­
Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:

  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 19/20 * 18/19 * 17/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
Show more
­
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,715
Own Kudos:
810,345
 [1]
Given Kudos: 105,795
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,715
Kudos: 810,345
 [1]
If 5 integers are to be selected at random and without replacement 14 Mar 2024, 02:14
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
unicornilove
Hi Bunuel, what is wrong with 1 - 18C5/20C5?

I take 18C5 to be choose 5 numbers when 10 and 20 are excluded. So in the equation 1 - 18C5/20C5 --> the solution gets us the probability when 10, 20 are included.

Thank you in advance for your kind response.
Bunuel

Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Show Spoiler
­
­
Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:


  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 19/20 * 18/19 * 17/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
­
Show more

­The opposite of choosing both 10 and 20 is not neither 10 nor 20. It's neither 10 nor 20, plus 10 but not 20, and plus 20 but not 10. So, in this case, it's better to stick with the direct approach.
User avatar
DanTheGMATMan
User avatar
Joined: 02 Oct 2015
Last visit: 18 Apr 2026
Posts: 380
Own Kudos:
267
 [4]
Given Kudos: 9
Expert
Expert reply
Posts: 380
Kudos: 267
 [4]
If 5 integers are to be selected at random and without replacement 19 Sep 2024, 09:43
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Just use the combination formula to count your total possible selections, and then the desired selections whereby you do select 10 and 20:

User avatar
hemanthPA
User avatar
Joined: 09 Mar 2023
Last visit: 18 Mar 2025
Posts: 47
Own Kudos:
42
 [10]
Given Kudos: 11
Posts: 47
Kudos: 42
 [10]
If 5 integers are to be selected at random and without replacement 06 Dec 2024, 02:45
6
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
Out of of 20 integers, 5 to be chosen

The probability that 10 occupies one of these 5 choices =1/20 + 1/20 + 1/20 + 1/20 + 1/20 = 5/20

19 integers remain, 4 slots left to fill.

The probability that 20 occupies one of these 4 slots = 1/19 + 1/19 + 1/19 + 1/19 = 4/19

Thus: P(both 10 and 20 are selected) = 5/20 * 4/19 = 20/20*19 = 1/19
User avatar
ravi1522
User avatar
Joined: 05 Jan 2023
Last visit: 21 Apr 2026
Posts: 171
Own Kudos:
111
Given Kudos: 5
Location: India
Concentration: General Management, General Management
Schools: INSEAD Jan '26
GMAT Focus 1: 595 Q80 V83 DI76
GMAT 1: 530 Q38 V24
GPA: 7.2
WE:Design (Real Estate)
Products:
GMAT Club Tests Forum Quiz
Schools: INSEAD Jan '26
GMAT Focus 1: 595 Q80 V83 DI76
GMAT 1: 530 Q38 V24
Posts: 171
Kudos: 111
If 5 integers are to be selected at random and without replacement 12 Dec 2024, 23:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello ,
5 integers are to be selected at random
Total number of ways in which 5 integers can be selected out of 20 integers is 20C5

Integers 10 and 20 to be there means remaining 3 integers to be selected out of 18 integers.
Total number of ways in which 3 integers can be selected out of 18 is 18C3*2C2

Required probability = 18C3*2C2/20C5
when you solve we get
1/19
User avatar
jacob46
User avatar
Joined: 02 Jan 2022
Last visit: 29 Aug 2025
Posts: 10
Own Kudos:
3
Given Kudos: 8
Posts: 10
Kudos: 3
If 5 integers are to be selected at random and without replacement 25 Mar 2025, 08:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sorry I couldn't get my head around the logic and I am confused on why the 2/20*1/19 would not work in this case, could you please help me understand the logic
Perhaps if you could quote a question where this works and 1/20*1/19 doesn't, it could really help.

Many thanks
Bunuel
kevinhirose
Hi Bunuel

Could you explain to me why we can’t do 2/20 * 1/19 * 1 * 1 * 1? Isn’t it 2/20 because we can either choose 10 or 20 and then it is 1/19 because after we choose 10 or 20 it’s whichever one is left?

I do understand how to do 18C3
/ 20C5 but I also get confused sometimes on problems like this and solve it wrong.

Posted from my mobile device
­The reason why using 2/20 for selecting either 10 or 20 and then 1/19 for selecting the remaining one is incorrect is because it results in double-counting certain outcomes.

For instance, selecting 10 first and then 20 is one outcome, and selecting 20 first and then 10 is another outcome. However, both outcomes essentially represent the same event: choosing both 10 and 20. By using 2/20, we're inadvertently counting this event twice.
Show more
User avatar
GMATGuruNY
User avatar
Joined: 04 Aug 2010
Last visit: 02 Apr 2026
Posts: 1,347
Own Kudos:
3,904
 [1]
Given Kudos: 9
Schools:Dartmouth College
Expert
Expert reply
Posts: 1,347
Kudos: 3,904
 [1]
If 5 integers are to be selected at random and without replacement 25 Mar 2025, 11:11
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jacob46
I am confused on why the 2/20*1/19 would not work in this case, could you please help me understand the logic.

Many thanks
kevinhiro
Show more
This approach is viable if only two numbers are being selected.
If two distinct numbers are selected from the first 20 positive integers, what is the probability that both 10 and 20 are selected?

P(first number is 10 or 20) = 2/20 (Of the 20 integers, two are 10 or 20)
P(second number is 10 or 20) = 1/19 (Of the 19 remaining integers, only one is 10 or 20)
Since we want both events to happen, we multiply the fractions:
2/20 * 1/19 = 1/190

Using combinatorics:
From 20 integers, the number of ways to choose 2 = 20C2 = \(\frac{20*19}{2*1} =190\)
Since there is only 1 good outcome -- choosing both 10 and 20 -- we get:
P(good outcome) = 1/190

If only two integers are selected, the probability that both 10 and 20 are chosen is quite small (1/190).
But in the problem at hand, five integers are selected.
Since in this case there are five opportunities to select 10 and 20 instead of only two, the probability that both 10 and 20 are among the selected integers is much greater than 1/190.
User avatar
SudsMeister
User avatar
Joined: 04 Sep 2019
Last visit: 29 Dec 2025
Posts: 86
Own Kudos:
34
Given Kudos: 97
Location: India
Concentration: Finance, Sustainability
Schools: LBS '22 (D) Instead Sept 20 Intake CBS '22 (D) Booth '22 (D)
GMAT Focus 1: 705 Q89 V85 DI81
GMAT 1: 750 Q50 V42
GPA: 79.3%
WE:Education (Energy)
Schools: LBS '22 (D) Instead Sept 20 Intake CBS '22 (D) Booth '22 (D)
GMAT Focus 1: 705 Q89 V85 DI81
GMAT 1: 750 Q50 V42
Posts: 86
Kudos: 34
If 5 integers are to be selected at random and without replacement 07 May 2025, 01:37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
einstein801
Hi Bunuel, what is wrong with 1 - 18C5/20C5?

I take 18C5 to be choose 5 numbers when 10 and 20 are excluded. So in the equation 1 - 18C5/20C5 --> the solution gets us the probability when 10, 20 are included.

Thank you in advance for your kind response.
Show more
If we use this approach then our answer will be:

Choosing both 10 and 20 = Total probability - choosing neither 10 nor 20 - choosing 10 but not 20 - choosing 20 but not 10

Total probability = 1

Choosing neither 10 nor 20 =\( \frac{18C5 }{ 20C5} = \frac{18*17*16*15*14}{5! }* \frac{ 5!}{20*19*18*17*16} =\frac{ 15*14 }{ 20*19} = \frac{21 }{ 38}\\
\)
Choosing 10 but not 20 (since 10 is chosen and 20 is not allowed, so 18 possible integers remain) = \(\frac{18C4 }{ 20C5 }= \frac{18*17*16*15}{4! } * \frac{ 5!}{20*19*18*17*16} = \frac{15*5}{20*19 }= \frac{15}{76}\)

Choosing 20 but no 10 = \(\frac{18C4 }{ 20C5} = \frac{15}{76}\)

Choosing both 10 and 20\( = 1 - \frac{ 21}{38 } - \frac{ 15}{76 } - \frac{ 15}{76} = \frac{(76 - 42 - 15 - 15)}{76 } = \frac{ 4}{76 } = \frac{ 1}{19}\)

Answer D.
User avatar
GMATinsight
User avatar
Major Poster
Joined: 08 Jul 2010
Last visit: 19 Apr 2026
Posts: 6,976
Own Kudos:
16,890
Given Kudos: 128
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Products:
My Reviews
Expert
Expert reply
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
Posts: 6,976
Kudos: 16,890
If 5 integers are to be selected at random and without replacement 07 May 2025, 01:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60
Show more
Total Outcomes = 20C5

Favorable outcomes = 3 more numbers out of 18 keeping 10 and 20 fixed = 18C3

Required Probability = 18C3 / 20C5 = 1/19

Answer: Option D

User avatar
Dragon805
User avatar
Joined: 04 Mar 2025
Last visit: 23 Feb 2026
Posts: 16
Own Kudos:
9
Given Kudos: 132
GMAT Focus 1: 695 Q84 V90 DI80
GMAT Focus 1: 695 Q84 V90 DI80
Posts: 16
Kudos: 9
If 5 integers are to be selected at random and without replacement 26 May 2025, 06:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Correct answer is option D.

Basically, we need to find out the probability of selecting 2 numbers from the set {10,20} and 3 numbers from the set of all numbers from 1 to 20 except for 10 and 20 (that is 18 numbers). The answer to this is same as the answer to what is the probability that 2 of the 5 selected integers will be 10 and 20.

number of ways in which 2 numbers can be selected from {10,20} = 2C2 = 1 (duh)
number of ways in which 3 numbers can be selected from the remaining 18 numbers = 18C3
number of ways in which you can select 5 numbers from the overall set of 20 numbers = 20C5

required probability = 18C3/20C5
if you know how to expand 18C3 and 20C5 then the calculation becomes very simple and it turns out to be
= [ 18*17*16 ]/[ 3*2*1 ] * [5*4*3*2*1]/[20*19*18*17*16]
everything cancels out and we get = 1/19
Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Show Spoiler
­
Show more
User avatar
HammeredDeepu97
User avatar
Joined: 01 Apr 2024
Last visit: 21 Apr 2026
Posts: 5
Own Kudos:
3
Posts: 5
Kudos: 3
If 5 integers are to be selected at random and without replacement 26 Jul 2025, 07:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey Bunuel,

Why is it 5! / 3! for distribution? The 3 numbers that we choose apart from 10 and 20, do we consider the (10&20) as a box and the rest of the 3 numbers as movable? Also why is the probability of choosing these numbers as 1?
Bunuel

Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:
  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 18/20 * 17/19 * 16/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:

\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,715
Own Kudos:
810,345
Given Kudos: 105,795
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,715
Kudos: 810,345
If 5 integers are to be selected at random and without replacement 27 Jul 2025, 04:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
HammeredDeepu97
Hey Bunuel,

Why is it 5! / 3! for distribution? The 3 numbers that we choose apart from 10 and 20, do we consider the (10&20) as a box and the rest of the 3 numbers as movable? Also why is the probability of choosing these numbers as 1?
Bunuel

Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:
  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 18/20 * 17/19 * 16/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:

\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
Show more

We multiply by 5!/3! because we're fixing two numbers (10 and 20) and choosing three others from the remaining 18. The 5!/3! accounts for the number of ways to arrange 10, 20, and the 3 other numbers within the 5 slots.

As for the "probability of choosing these numbers is 1" part, once 10 and 20 are chosen, the remaining 3 can be any numbers from the other 18, so each of those has probability 1. The randomness is only in choosing 10 and 20; the rest are unrestricted.
User avatar
Isha2701
User avatar
Joined: 28 Dec 2024
Last visit: 21 Apr 2026
Posts: 9
Given Kudos: 419
Products:
Forum Quiz
Posts: 9
Kudos: 0
If 5 integers are to be selected at random and without replacement 17 Apr 2026, 23:20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel - what does without replacement mean here? I understand that we choose 10 and 20 def, meaning 1/20 x 1/19 x 18/18 x 17/17 x 16/16 (then for rest three numbers bcause we have 18 left now and any 18 can be chosen, similarly for 17 and 16 left.) And they all can be arranged in 5! ways..so where am I wrong?
 1   2   
Moderators:
Bunuel
Math Expert
109715 posts
Krunaal
Tuck School Moderator
853 posts