Last visit was: 19 Nov 2025, 07:47 It is currently 19 Nov 2025, 07:47
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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
User avatar
LalaB
User avatar
Current Student
Joined: 23 Oct 2010
Last visit: 17 Jul 2016
Posts: 227
Own Kudos:
1,327
 [95]
Given Kudos: 73
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Posts: 227
Kudos: 1,327
 [95]
The integers n and t are positive and n > t > 1. How many 31 Mar 2012, 07:34
10
Kudos
Add Kudos
85
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:

75% (hard)

Question Stats:

51% (01:52) correct 49% (02:07) wrong based on 1213 sessions

History

Date
Time
Result
Not Attempted Yet
The integers n and t are positive and n > t > 1. How many different subgroups of t items can be formed from a group of n different items?

(1) The number of different subgroups of n − t different items that can be formed from a group of n different items is 680.

(2) nt = 51

good kaplan question. just wanted to share with you.
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,253
 [33]
Given Kudos: 99,977
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: 105,389
Kudos: 778,253
 [33]
The integers n and t are positive and n > t > 1. How many 31 Mar 2012, 07:49
10
Kudos
Add Kudos
22
Bookmarks
Bookmark this Post
The integers n and t are positive and n > t > 1. How many different subgroups of t items can be formed from a group of n different items?

The question basically asks about the value of \(C^t_n=\frac{n!}{(n-t)!*t!}\).

(1) The number of different subgroups of n − t different items that can be formed from a group of n different items is 680 --> \(C^{n-t}_n=\frac{n!}{(n-(n-t))!*(n-t)!}=\frac{n!}{t!*(n-t)!}=680\), directly gives us the asnwer. Sufficient.

(2) nt = 51 --> 51=17*3=51*1, since n > t > 1 then t=3 and n=17 --> \(C^t_n=\frac{n!}{(n-t)!*t!}=\frac{17!}{14!*3!}=680\). Sufficient.

Answer; D.
General Discussion
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,253
 [3]
Given Kudos: 99,977
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: 105,389
Kudos: 778,253
 [3]
The integers n and t are positive and n > t > 1. How many 01 Jul 2013, 00:46
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html
User avatar
ziko
User avatar
Joined: 28 Feb 2012
Last visit: 29 Jan 2014
Posts: 91
Own Kudos:
217
 [3]
Given Kudos: 17
Concentration: Strategy, International Business
Schools: INSEAD Jan '13
GPA: 3.9
WE:Marketing (Other)
Schools: INSEAD Jan '13
Posts: 91
Kudos: 217
 [3]
The integers n and t are positive and n > t > 1. How many 01 Jul 2013, 01:18
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
LalaB
The integers n and t are positive and n > t > 1. How many different subgroups of t items can be formed from a group of n different items?

(1) The number of different subgroups of n − t different items that can be formed from a group of n different items is 680.

(2) nt = 51

good kaplan question. just wanted to share with you.
Show more


One of the easier ways to solve DS questions is always to look at each answer choice at glance. Just to find whether there is any idea comes to mind by looking one or another.
For example in this problem i have looked at the 1) statement and had a doubt, then i did not spend much time to solve it and moved to the 2).

2) statement tells us that these integers must be n=17 and t=3, considering the conditions of the question there are no other choices. So we can solve this problem. Although we do not need to it i will solve it: we are looking for how many groups of 3 integers we can form out of 17 (considering that 1 would not make any difference i am saying 17).
(17*16*15)/3*2=680

By looking at this answer choice we can state that it is the same as the answer which is given in the statement (1). You can try couple of different values to get the combination of 680, but i am more than sure that you will not be able to get such combination without using 17 and 3.
So i am concluding that the answer is D.
User avatar
SunthoshiTejaswi
User avatar
Joined: 05 Feb 2014
Last visit: 24 Jan 2018
Posts: 14
Given Kudos: 18
Location: India
Concentration: Human Resources, General Management
Schools: Tepper '20 (S)
GMAT 1: 720 Q49 V40
GPA: 3.33
Schools: Tepper '20 (S)
GMAT 1: 720 Q49 V40
Posts: 14
Kudos: 0
The integers n and t are positive and n > t > 1. How many 14 Sep 2016, 07:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
The integers n and t are positive and n > t > 1. How many different subgroups of t items can be formed from a group of n different items?

The question basically asks about the value of \(C^t_n=\frac{n!}{(n-t)!*t!}\).

(1) The number of different subgroups of n − t different items that can be formed from a group of n different items is 680 --> \(C^{n-t}_n=\frac{n!}{(n-(n-t))!*(n-t)!}=\frac{n!}{t!*(n-t)!}=680\), directly gives us the asnwer. Sufficient.

(2) nt = 51 --> 51=17*3=51*1, since n > t > 1 then t=3 and n=17 --> \(C^t_n=\frac{n!}{(n-t)!*t!}=\frac{17!}{14!*3!}=680\). Sufficient.

Answer; D.
Show more

Hi Bunnuel

Can we consider this as a ratio problem where we are asked to find the ratio of n to t. Even by this logic both the statements are enough!!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,253
Given Kudos: 99,977
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: 105,389
Kudos: 778,253
The integers n and t are positive and n > t > 1. How many 14 Sep 2016, 08:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
SunthoshiTejaswi
Bunuel
The integers n and t are positive and n > t > 1. How many different subgroups of t items can be formed from a group of n different items?

The question basically asks about the value of \(C^t_n=\frac{n!}{(n-t)!*t!}\).

(1) The number of different subgroups of n − t different items that can be formed from a group of n different items is 680 --> \(C^{n-t}_n=\frac{n!}{(n-(n-t))!*(n-t)!}=\frac{n!}{t!*(n-t)!}=680\), directly gives us the asnwer. Sufficient.

(2) nt = 51 --> 51=17*3=51*1, since n > t > 1 then t=3 and n=17 --> \(C^t_n=\frac{n!}{(n-t)!*t!}=\frac{17!}{14!*3!}=680\). Sufficient.

Answer; D.

Hi Bunnuel

Can we consider this as a ratio problem where we are asked to find the ratio of n to t. Even by this logic both the statements are enough!!
Show more

Just knowing the ratio won't be enough. For example, knowing that n/t = 3/2, won't be enough to answer the question.
avatar
cyd37
avatar
Joined: 26 May 2018
Last visit: 07 Feb 2021
Posts: 3
Given Kudos: 294
Posts: 3
Kudos: 0
The integers n and t are positive and n > t > 1. How many 06 Sep 2018, 04:15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel,

How did you simplify the (n-(n-t!)) * t!? I dont understand how you made it n-t! * t!

Thanks

Bunuel
The integers n and t are positive and n > t > 1. How many different subgroups of t items can be formed from a group of n different items?

The question basically asks about the value of \(C^t_n=\frac{n!}{(n-t)!*t!}\).

(1) The number of different subgroups of n − t different items that can be formed from a group of n different items is 680 --> \(C^{n-t}_n=\frac{n!}{(n-(n-t))!*(n-t)!}=\frac{n!}{t!*(n-t)!}=680\), directly gives us the asnwer. Sufficient.

(2) nt = 51 --> 51=17*3=51*1, since n > t > 1 then t=3 and n=17 --> \(C^t_n=\frac{n!}{(n-t)!*t!}=\frac{17!}{14!*3!}=680\). Sufficient.

Answer; D.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,253
 [1]
Given Kudos: 99,977
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: 105,389
Kudos: 778,253
 [1]
The integers n and t are positive and n > t > 1. How many 06 Sep 2018, 04:19
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
cyd37
Hi Bunuel,

How did you simplify the (n-(n-t!)) * t!? I dont understand how you made it n-t! * t!

Thanks

Bunuel
The integers n and t are positive and n > t > 1. How many different subgroups of t items can be formed from a group of n different items?

The question basically asks about the value of \(C^t_n=\frac{n!}{(n-t)!*t!}\).

(1) The number of different subgroups of n − t different items that can be formed from a group of n different items is 680 --> \(C^{n-t}_n=\frac{n!}{(n-(n-t))!*(n-t)!}=\frac{n!}{t!*(n-t)!}=680\), directly gives us the asnwer. Sufficient.

(2) nt = 51 --> 51=17*3=51*1, since n > t > 1 then t=3 and n=17 --> \(C^t_n=\frac{n!}{(n-t)!*t!}=\frac{17!}{14!*3!}=680\). Sufficient.

Answer; D.
Show more


\((n-(n-t))!*(n-t)! = (n-n+t)!*(n-t)! = t!*(n-t)!\)
User avatar
CorporateAsset
User avatar
Joined: 25 Aug 2024
Last visit: 23 Jul 2025
Posts: 48
Own Kudos:
32
 [1]
Given Kudos: 255
Location: India
Concentration: Entrepreneurship, Strategy
GPA: 4
Products:
My Reviews
Posts: 48
Kudos: 32
 [1]
The integers n and t are positive and n > t > 1. How many 13 Jun 2025, 03:04
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
For statement A:
We’re told that the number of subgroups of size n−t from n items is 680.
At first, this might seem unrelated because we want the number of subgroups of size t, not n−t. But here’s the key insight:
In combinations, choosing t items from n is the same as choosing the n−t items you’re not selecting. This is known as the symmetry rule of combinations:

So if we know there are 680 ways to choose n−t items from n, then there are also 680 ways to choose t items from n. That’s exactly what the question is asking.
Moderators:
Bunuel
Math Expert
105389 posts
kingbucky
496 posts