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

 It is currently 23 May 2019, 00:12

### 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

# How many unique quadrilaterals can be inscribed in the vertices of a

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55236
How many unique quadrilaterals can be inscribed in the vertices of a  [#permalink]

### Show Tags

19 Jul 2017, 23:24
00:00

Difficulty:

55% (hard)

Question Stats:

52% (02:10) correct 48% (02:31) wrong based on 47 sessions

### HideShow timer Statistics

How many unique quadrilaterals can be inscribed in the vertices of a nonagon (a 9-sided figure), if points A and B, two vertices in the nonagon, cannot make up the same quadrilateral?

(A) 126

(B) 105

(C) 96

(D) 65

(E) 21

_________________
Intern
Joined: 12 Dec 2010
Posts: 42
Re: How many unique quadrilaterals can be inscribed in the vertices of a  [#permalink]

### Show Tags

Updated on: 20 Jul 2017, 03:56
IMHO B

Total number of quadrilaterals that can be formed from a nonagon is 9C4 = 126.

Number of quadrilaterals that can be formed with 2 points A and B in the same quadrilateral is 2C2 * 7C2 = 21

Now number of quadrilaterals with A and B not together is 126 - 21= 105

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

Originally posted by mcmoorthy on 20 Jul 2017, 01:56.
Last edited by mcmoorthy on 20 Jul 2017, 03:56, edited 1 time in total.
Manager
Joined: 02 Nov 2015
Posts: 163
GMAT 1: 640 Q49 V29
Re: How many unique quadrilaterals can be inscribed in the vertices of a  [#permalink]

### Show Tags

20 Jul 2017, 03:39
mcmoorthy wrote:
IMHO B

Total number of quadrilaterals that can be formed from a nonagon is 9C4 = 126.

Number of quadrilaterals that can be formed with 2 points A and B in the same quadrilateral is 2C2 * 8C2 = 21

Now number of quadrilaterals with A and B not together is 126 - 21= 105

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

Hi Murthy,

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app
Intern
Joined: 05 Nov 2016
Posts: 11
Re: How many unique quadrilaterals can be inscribed in the vertices of a  [#permalink]

### Show Tags

20 Jul 2017, 03:48
mcmoorthy wrote:
IMHO B

Total number of quadrilaterals that can be formed from a nonagon is 9C4 = 126.

Number of quadrilaterals that can be formed with 2 points A and B in the same quadrilateral is 2C2 * 8C2 = 21

Now number of quadrilaterals with A and B not together is 126 - 21= 105

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

I think you've made a small mistake over there, it should be 7c2 since you have already selected 2 points and you will have to select other 2 points from remaining 7 points.

Sent from my Redmi Note 3 using GMAT Club Forum mobile app
Intern
Joined: 12 Dec 2010
Posts: 42
Re: How many unique quadrilaterals can be inscribed in the vertices of a  [#permalink]

### Show Tags

20 Jul 2017, 03:57
Edited. I had written 8 instead of 7. The number is 21 anyway. Thanks for the nudge.

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2823
Re: How many unique quadrilaterals can be inscribed in the vertices of a  [#permalink]

### Show Tags

25 Jul 2017, 11:24
Bunuel wrote:
How many unique quadrilaterals can be inscribed in the vertices of a nonagon (a 9-sided figure), if points A and B, two vertices in the nonagon, cannot make up the same quadrilateral?

(A) 126

(B) 105

(C) 96

(D) 65

(E) 21

Since a nonagon has 9 vertices and a quadrilateral has 4 vertices, the number of quadrilaterals that can be made is 9C4 = (9 x 8 x 7 x 6)/4! = (9 x 8 x 7 x 6)/(4 x 3 x 2) = 3 x 7 x 6 = 126, if there are no restrictions. However, since vertices A and B can’t both be in the same quadrilateral, we need to subtract the number of quadrilaterals that have both vertices A and B. The number of such quadrilaterals is 2C2 x 7C2 = 1 x (7 x 6)/2! = 42/2 = 21 (notice that 2C2 is the number of ways A and B can be picked if they have to be 2 vertices of the quadrilateral and 7C2 is the number of ways the other 2 vertices of the quadrilateral can be picked from the other 7 vertices).

Thus, the number of quadrilaterals such that vertices A and B are not in the same quadrilateral is 126 - 21 = 105.

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: How many unique quadrilaterals can be inscribed in the vertices of a   [#permalink] 25 Jul 2017, 11:24
Display posts from previous: Sort by