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

 It is currently 18 Oct 2019, 12:23

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

In a triangle ABC, there are 5, 6 and 4 points, different from vertice

Author Message
TAGS:

Hide Tags

Status: Preparing for GMAT
Joined: 25 Nov 2015
Posts: 1043
Location: India
GPA: 3.64
In a triangle ABC, there are 5, 6 and 4 points, different from vertice  [#permalink]

Show Tags

21 Jan 2018, 11:23
00:00

Difficulty:

95% (hard)

Question Stats:

39% (03:31) correct 61% (02:27) wrong based on 48 sessions

HideShow timer Statistics

In a triangle ABC, there are 5, 6 and 4 points, different from the vertices A, B and C of the triangle, on the three sides AB, BC and CA respectively. How many distinct triangles can be drawn using the 18 points (including the vertices A, B and C)?
A) 111
B) 705
C) 816
D) 910
E) 955

_________________
Please give kudos, if you like my post

When the going gets tough, the tough gets going...
examPAL Representative
Joined: 07 Dec 2017
Posts: 1153
Re: In a triangle ABC, there are 5, 6 and 4 points, different from vertice  [#permalink]

Show Tags

21 Jan 2018, 11:50
2
souvonik2k wrote:
In a triangle ABC, there are 5, 6 and 4 points, different from the vertices A, B and C of the triangle, on the three sides AB, BC and CA respectively. How many distinct triangles can be drawn using the 18 points (including the vertices A, B and C)?
A) 111
B) 705
C) 816
D) 910
E) 955

As direct calculation involves many possible cases, we'll solve this by calculating "1 - the probability".
This is a Logical approach.

There are 18C3 ways to choose 3 vertices from 18 points.
This is 18!/(15!*3!) = (18*17*16)/(3*2*1) = 3*17*16 = 51*16 = 510+306=816
(C),(D),(E) are eliminated as they are too large (we know we're going to have less than 816).
And since (A) looks way too small, we should feel comfortable guessing (B).

Let's calculate: of all the ways to choose vertices, the only 'illegal' choice is when all 3 points are on the same side.
All 3 on AB (including points A,B): 7C3 = (7*6*5)/(3*2*1) = 35
All 3 on BC (including points B,C): 8C3 = (8*7*6)/(3*2*1) = 56
All 3 on AB (including points A,B): 6C3 = (6*5*4)/(3*2*1) = 20

35+56+20 =111
816 - 111 = 705
_________________
Re: In a triangle ABC, there are 5, 6 and 4 points, different from vertice   [#permalink] 21 Jan 2018, 11:50
Display posts from previous: Sort by