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Let T_n be the number of all possible triangles formed by

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Let T_n be the number of all possible triangles formed by  [#permalink]

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New post Updated on: 26 Mar 2014, 07:19
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Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is

A. 5
B. 6
C. 7
D. 8
E. 10

Originally posted by arunspanda on 26 Mar 2014, 06:17.
Last edited by Bunuel on 26 Mar 2014, 07:19, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Let T_n be the number of all possible triangles formed by  [#permalink]

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New post 26 Mar 2014, 07:21
arunspanda wrote:
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is

A. 5
B. 6
C. 7
D. 8
E. 10


In a plane if there are n points of which no three are collinear, then the number of triangles that can be formed by joining them is \(C^3_n\).

We are given that \(T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10\) --> \(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\) --> \(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\) --> \((n-1)n=20\) --> \(n=5\).

Answer: A.

Or: one additional point gives 10 more triangles, so when we add one point there are 10 different pairs of points which make triangles with that additional point --> \(C^2_n=10\) --> \((n-1)n=20\) --> \(n=5\).

Answer: A.

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Re: Let T_n be the number of all possible triangles formed by  [#permalink]

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New post 26 Feb 2019, 07:44
Bunuel wrote:
arunspanda wrote:
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is

A. 5
B. 6
C. 7
D. 8
E. 10


In a plane if there are n points of which no three are collinear, then the number of triangles that can be formed by joining them is \(C^3_n\).

We are given that \(T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10\) --> \(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\) -->\(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\) --> \((n-1)n=20\)--> \(n=5\).

Answer: A.

Or: one additional point gives 10 more triangles, so when we add one point there are 10 different pairs of points which make triangles with that additional point --> \(C^2_n=10\) --> \((n-1)n=20\) --> \(n=5\).

Answer: A.


Bunuel can you please expand upon the highlighted part? I get that we are cross multiplying for LCD, but I am getting lost.

Also, is there any way this is related to n-2 (number of triangles from a regular polygon)? I tried to use the idea that n(n-2) would give us the number of vertices*number of triangles formed from each.

Tn = n(n-2)
Tn+1 = n(n-2) + 10
... ?
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Re: Let T_n be the number of all possible triangles formed by  [#permalink]

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New post 26 Feb 2019, 08:07
energetics wrote:
Bunuel wrote:
arunspanda wrote:
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is

A. 5
B. 6
C. 7
D. 8
E. 10


In a plane if there are n points of which no three are collinear, then the number of triangles that can be formed by joining them is \(C^3_n\).

We are given that \(T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10\) --> \(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\) -->\(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\) --> \((n-1)n=20\)--> \(n=5\).

Answer: A.

Or: one additional point gives 10 more triangles, so when we add one point there are 10 different pairs of points which make triangles with that additional point --> \(C^2_n=10\) --> \((n-1)n=20\) --> \(n=5\).

Answer: A.


Bunuel can you please expand upon the highlighted part? I get that we are cross multiplying for LCD, but I am getting lost.

Also, is there any way this is related to n-2 (number of triangles from a regular polygon)? I tried to use the idea that n(n-2) would give us the number of vertices*number of triangles formed from each.

Tn = n(n-2)
Tn+1 = n(n-2) + 10
... ?



\(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\);

Express (n+1)! as (n-2)!*(n-1)n(n+1) and n! as (n-3)!*(n-2)(n-1)n:

\(\frac{(n-2)!*(n-1)n(n+1)}{3!(n-2)!}-\frac{(n-3)!*(n-2)(n-1)n}{3!(n-3)!}=10\)

\(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\)

\(\frac{(n-1)n(n+1)-(n-2)(n-1)n}{6}=10\)

\(\frac{(n-1)n(n+1-n+2)}{6}=10\)

\(\frac{(n-1)n*3}{6}=10\)

\((n-1)n=20\)

\(n=5\).
_________________
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Resources:
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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: Let T_n be the number of all possible triangles formed by  [#permalink]

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New post 26 Feb 2019, 11:08
We could also solve this by directly inputting the numbers. 6c3-5c3=10.Hence value of n is 5

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Re: Let T_n be the number of all possible triangles formed by   [#permalink] 26 Feb 2019, 11:08
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