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The integers n and t are positive and n > t > 1. How many

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The integers n and t are positive and n > t > 1. How many  [#permalink]

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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.

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Re: The integers n and t are positive and n > t > 1. How many  [#permalink]

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New post 31 Mar 2012, 06:49
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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.
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Re: The integers n and t are positive and n > t > 1. How many  [#permalink]

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New post 30 Jun 2013, 23:46
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: The integers n and t are positive and n > t > 1. How many  [#permalink]

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New post 01 Jul 2013, 00:18
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LalaB wrote:
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.



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.
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Re: The integers n and t are positive and n > t > 1. How many  [#permalink]

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New post 14 Sep 2016, 06:22
Bunuel wrote:
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!!
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Re: The integers n and t are positive and n > t > 1. How many  [#permalink]

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New post 14 Sep 2016, 07:19
SunthoshiTejaswi wrote:
Bunuel wrote:
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!!


Just knowing the ratio won't be enough. For example, knowing that n/t = 3/2, won't be enough to answer the question.
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Collection of Questions:
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Re: The integers n and t are positive and n > t > 1. How many  [#permalink]

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New post 06 Sep 2018, 03:15
Hi Bunuel,

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

Thanks

Bunuel wrote:
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.
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Re: The integers n and t are positive and n > t > 1. How many  [#permalink]

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New post 06 Sep 2018, 03:19
cyd37 wrote:
Hi Bunuel,

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

Thanks

Bunuel wrote:
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.



\((n-(n-t))!*(n-t)! = (n-n+t)!*(n-t)! = t!*(n-t)!\)
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New to the Math Forum?
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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The integers n and t are positive and n > t > 1. How many &nbs [#permalink] 06 Sep 2018, 03:19
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