January 20, 2019 January 20, 2019 07:00 AM PST 07:00 AM PST Get personalized insights on how to achieve your Target Quant Score. January 21, 2019 January 21, 2019 10:00 PM PST 11:00 PM PST Mark your calendars  All GMAT Club Tests are free and open January 21st for celebrate Martin Luther King Jr.'s Birthday.
Author 
Message 
Intern
Joined: 30 Aug 2010
Posts: 13

How many diagonals does a polygon with 21 sides have, if one
[#permalink]
Show Tags
23 Sep 2010, 01:59
Question Stats:
64% (01:11) correct 36% (01:16) wrong based on 296 sessions
HideShow timer Statistics
How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal? A. 21 B. 170 C. 340 D. 357 E. 420 I was considering a different approach of 20 vertices (21 vertices 1 vertex not linking) linking to 17 (21 vertices  1 vertix I'm considering 2 vertices next to the one I'm considering  1 vertex not linking) others each: that is 17*20=340. We divide that by 2 since every diagonal connects two vertices. 340/2=170.
Consider the parallel case in which we have 21 people who shake hands to each other. In that case we'll have \(21*20/2\) handshakes. This because, if we consider all the 21 people in a line, the first one will shake hands to 20, the second to 19, the third to 18 people and so on. So we have: 20+19+18+...+1 This sum equals to \(21*20/2=210\).
Handshaking and diagonals are different in counting since diagonals don't include sides. To calculate diagonals we have to subtract the number of sides (that are 21) from \(21*20/2\). So: N° of diagonals = \((21*20/2)21=21021=189\)
Since one vertex does not connect to any diagonals, we have to subtract the diagonals of this vertex. Diagonals of one vertex equals to: 21 (number of vertex)  1 (itself)  2 (sides with vertices next to it).
So the answer should be: \(18918=171\) That's different from 170 obtained above.
What's wrong with my second approach?
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 52296

Re: How many diagonals does a polygon with 21 sides have...
[#permalink]
Show Tags
23 Sep 2010, 02:49



Retired Moderator
Joined: 02 Sep 2010
Posts: 765
Location: London

Re: How many diagonals does a polygon with 21 sides have...
[#permalink]
Show Tags
23 Sep 2010, 03:04
Take any vertex in the polygon. To form a diagnol, it can be connected to any one of 18 vertices (counting out itself, and the 2 vertices adjacent to it). If we count diagnols like this, we will double count everything (once for every end point). So the total number of diagnols = (18*21)/2 = 189 Each vertex has 18 diagnols, so if we take all the diagnols from this one vertex out .. we are left with 171 diagnolsAlternate SolutionConsider a polygon with 20 sides. By logic similar to above, the number of diagnols is (17*20)/2 = 170 Now take a new point X and any side of the 20sided figure AB. Join XA and XB to form a 21 sided figure. In this new figure all the old diagnols are still diagnols + the side AB also becomes a diagnol. Also note that there are no diagnols originating from X Hence the number of diagnols = 170 + 1 (AB) = 171 diagnols
_________________
Math writeups 1) Algebra101 2) Sequences 3) Set combinatorics 4) 3D geometry
My GMAT story
GMAT Club Premium Membership  big benefits and savings



Math Expert
Joined: 02 Sep 2009
Posts: 52296

Re: How many diagonals does a polygon with 21 sides have...
[#permalink]
Show Tags
23 Sep 2010, 03:05
rraggio wrote: I was considering a different approach of 20 vertices (21 vertices 1 vertex not linking) linking to 17 (21 vertices  1 vertix I'm considering 2 vertices next to the one I'm considering  1 vertex not linking) others each: that is 17*20=340. We divide that by 2 since every diagonal connects two vertices. 340/2=170.
Consider the parallel case in which we have 21 people who shake hands to each other. In that case we'll have \(21*20/2\) handshakes. This because, if we consider all the 21 people in a line, the first one will shake hands to 20, the second to 19, the third to 18 people and so on. So we have: 20+19+18+...+1 This sum equals to \(21*20/2=210\).
Handshaking and diagonals are different in counting since diagonals don't include sides. To calculate diagonals we have to subtract the number of sides (that are 21) from \(21*20/2\). So: N° of diagonals = \((21*20/2)21=21021=189\)
Since one vertex does not connect to any diagonals, we have to subtract the diagonals of this vertex. Diagonals of one vertex equals to: 21 (number of vertex)  1 (itself)  2 (sides with vertices next to it).
So the answer should be: \(18918=171\) That's different from 170 obtained above.
What's wrong with my second approach? Actually first approach is wrong: 17*20 is not right. As 2 vertices which are next to excluded vertex can be connected each to 213=18 vertices, so 2*18. Other 18 vertices can be connected to 21  3  1 (excluded vertex) = 17 vertices. So we would have 2*18+18*17=342 > divided by 2 to exclude double counting > 342/2=171.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! 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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2591
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: How many diagonals does a polygon with 21 sides have...
[#permalink]
Show Tags
23 Sep 2010, 05:47
shrouded1 wrote: Consider a polygon with 20 sides. By logic similar to above, the number of diagnols is (17*20)/2 = 170
Now take a new point X and any side of the 20sided figure AB. Join XA and XB to form a 21 sided figure. In this new figure all the old diagnols are still diagnols + the side AB also becomes a diagnol. Also note that there are no diagnols originating from X
Hence the number of diagnols = 170 + 1 (AB) = 171 diagnols I did same way and got 170 forgot to add +1. Thanks !!
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Intern
Joined: 11 Sep 2010
Posts: 3

Polygon problems a sucking combinationHELP!
[#permalink]
Show Tags
10 Nov 2010, 05:59
Hey Guys, I need some help out here...I picked up this question from a list of questions posted by a fellow gmat geek. I am not able to figure out out how to solve this question in terms of whether I should I directly work out a 20c2 for this question.
To sum it up please let me know how to solve problems like these in terms of what should be the basic approach.
How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?
21 170 340 357 420



Intern
Joined: 11 Sep 2010
Posts: 3

Re: How many diagonals does a polygon with 21 sides have...
[#permalink]
Show Tags
10 Nov 2010, 06:32
Thanks guys! I appreciate your quick response!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8795
Location: Pune, India

Re: How many diagonals does a polygon with 21 sides have...
[#permalink]
Show Tags
10 Nov 2010, 07:17
It is good to remember that a polygon with n sides has nC2  n diagonals and follow up from there as Bunuel did or follow shrouded1's approach, which I thought was pretty cool too. Though, remember that if you are lost with the 21 sides, cannot think of the formula and are hard pressed for time, try and work it on a smaller scale. Lets see what happens in a 6 sided figure which is extremely easy to draw: 6 sided figure. 1 vertex left alone. When you start from the first point, you cant join it to 3 of the 6 points  itself, point left alone and point next to it. Total diagonals drawn: 3 + 2 + 1 Attachment:
Ques.jpg [ 18.05 KiB  Viewed 16540 times ]
21 sided figure. When you will start with the first point, you will not join it to 3 of the 21 points  itself, point left alone and point next to it. Total diagonals drawn will be: 18 + 17 + 16 + ... +1 = 18*19/2 = 171
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 04 Sep 2012
Posts: 3

Re: How many diagonals does a polygon with 21 sides have...
[#permalink]
Show Tags
10 Dec 2013, 19:46
Bunuel wrote: rraggio wrote: How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?
a) 21 b) 170 c) 340 d) 357 e) 420 I don't think B (170) is a correct answer. Generally the # of diagonals in \(n\) sided polygon equals to \(C^2_nn=\frac{n(n3)}{2}\): \(C^2_n\) choose any two vertices out of \(n\) to connect minus \(n\) sides, which won't be diagonals. So, # of diagonals in 21 sided polygon is \(C^2_nn=21021=189\). Since the diagonals from 1 particulat vertex shouldn't be counted then \(189(213)=171\) (one vertex makes \(n3\) diagonals). Answer: 171. Why don't we do simple calculation? Let's say there are 21 sides > there are total 19 "dots" to connect each diagonals So, 19 "dots" and 2 points to connect for each other = 2C19 = 171



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8795
Location: Pune, India

Re: How many diagonals does a polygon with 21 sides have...
[#permalink]
Show Tags
10 Dec 2013, 20:15
ltkenny wrote: Why don't we do simple calculation? Let's say there are 21 sides > there are total 19 "dots" to connect each diagonals So, 19 "dots" and 2 points to connect for each other = 2C19 = 171
How do you have 19 dots? If it is a 21 sided figure, it will have 21 vertices. One vertex is not participating in diagonals so you have 20 vertices participating. You select 2 out of these 20 in 20C2 ways. But this includes the 19 sides of the polygon (since one vertex is not participating, 2 sides are already not formed so you only need to deduct 19 sides.) So total no of diagonals will be 20C2  19 = 171
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 15 Mar 2012
Posts: 44

Re: How many diagonals does a polygon with 21 sides have, if one
[#permalink]
Show Tags
26 Jun 2014, 02:22
Bunuel wrote: rraggio wrote: How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?
a) 21 b) 170 c) 340 d) 357 e) 420 I don't think B (170) is a correct answer. Generally the # of diagonals in \(n\) sided polygon equals to \(C^2_nn=\frac{n(n3)}{2}\): \(C^2_n\) choose any two vertices out of \(n\) to connect minus \(n\) sides, which won't be diagonals. So, # of diagonals in 21 sided polygon is \(C^2_nn=21021=189\). Since the diagonals from 1 particulat vertex shouldn't be counted then \(189(213)=171\) (one vertex makes \(n3\) diagonals). Answer: 171. Q: How many diagonals does a polygon with 18 sides have if three of its vertices do not send any diagonal? A: #lines with (183) vertices = 15C2 = 15 x 14 / 2 = 105 3 vertices would form 4 sides => 4 sides not counted in 105 already Therefore, #diagonals = #lines  #sides = 105  (18  4) = 91 > (1) Is 91 correct? If I calculate it using the formulae, #diagonals = n (n3)/2 & Each vertex sends of n3 diagonals #desired diagonals = #diagonals with 18 sides  #diagonals 3 vertices would send = 18 (183) / 2  (183) * 3 = 135  45 = 90 > (2) Could you please tell why I see this difference in the answers (1) and (2)? Thank you very much in advance.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8795
Location: Pune, India

Re: How many diagonals does a polygon with 21 sides have, if one
[#permalink]
Show Tags
29 Jun 2014, 23:21
divineacclivity wrote: Q: How many diagonals does a polygon with 18 sides have if three of its vertices do not send any diagonal? A: #lines with (183) vertices = 15C2 = 15 x 14 / 2 = 105 3 vertices would form 4 sides => 4 sides not counted in 105 already Therefore, #diagonals = #lines  #sides = 105  (18  4) = 91 > (1) Is 91 correct?
If I calculate it using the formulae, #diagonals = n (n3)/2 & Each vertex sends of n3 diagonals #desired diagonals = #diagonals with 18 sides  #diagonals 3 vertices would send = 18 (183) / 2  (183) * 3 = 135  45 = 90 > (2)
Could you please tell why I see this difference in the answers (1) and (2)?
Thank you very much in advance.
Yes, 91 is correct. The second method is not correct because when you subtract each of the 15 diagonals made by the 3 vertices which we need to ignore, you are double counting 1 diagonal. You actually need to subtract only 44 diagonals. Which diagonal are you double counting? The one which connects 2 of the three ignored vertices. Try to make a polygon with a few vertices. Make a few diagonals. Remove 3 vertices next to each other. 2 of the three vertices which have a vertex between them will be joined by a diagonal. When you remove 15 diagonals for each vertex, you are removing that diagonal twice. Hence, what you need to do is 135  44 = 91. Mind you, we assumed that the vertices which were removed were next to each other. If they are not, the answer would be different since there would be more double counting.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 26 Aug 2014
Posts: 1

Re: How many diagonals does a polygon with 21 sides have, if one
[#permalink]
Show Tags
26 Aug 2014, 11:00
The best answer is B. We have 20 vertices(1 vertex is not working) linking to 17 others(1 vertex will make 17 diagonals) each: that is 17*20=340. We divide that by 2 since every diagonal connects two vertices. 340/2=170. The vertex that does not connect to any diagonal is just not counted



NonHuman User
Joined: 09 Sep 2013
Posts: 9454

Re: How many diagonals does a polygon with 21 sides have, if one
[#permalink]
Show Tags
14 Jul 2018, 18:40
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: How many diagonals does a polygon with 21 sides have, if one &nbs
[#permalink]
14 Jul 2018, 18:40






