GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 25 Mar 2019, 19:20

### 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

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

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 04 Oct 2013
Posts: 152
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Let T_n be the number of all possible triangles formed by  [#permalink]

### Show Tags

Updated on: 26 Mar 2014, 07:19
2
11
00:00

Difficulty:

85% (hard)

Question Stats:

50% (02:25) correct 50% (02:15) wrong based on 148 sessions

### HideShow timer Statistics

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.
Math Expert
Joined: 02 Sep 2009
Posts: 53831
Re: Let T_n be the number of all possible triangles formed by  [#permalink]

### Show Tags

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

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 --> $$C^2_n=10$$ --> $$(n-1)n=20$$ --> $$n=5$$.

_________________
Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 141
Re: Let T_n be the number of all possible triangles formed by  [#permalink]

### Show Tags

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

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 --> $$C^2_n=10$$ --> $$(n-1)n=20$$ --> $$n=5$$.

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
... ?
Math Expert
Joined: 02 Sep 2009
Posts: 53831
Re: Let T_n be the number of all possible triangles formed by  [#permalink]

### Show Tags

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

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 --> $$C^2_n=10$$ --> $$(n-1)n=20$$ --> $$n=5$$.

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:

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

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

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

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

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

$$(n-1)n=20$$

$$n=5$$.
_________________
Intern
Joined: 17 May 2018
Posts: 44
Re: Let T_n be the number of all possible triangles formed by  [#permalink]

### Show Tags

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

Posted from my mobile device
Re: Let T_n be the number of all possible triangles formed by   [#permalink] 26 Feb 2019, 11:08
Display posts from previous: Sort by