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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 = \((21*20/2)-21=210-21=189\)

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: \(189-18=171\) That's different from 170 obtained above.

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 \(C^2_n-n=\frac{n(n-3)}{2}\): \(C^2_n\) 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 \(C^2_n-n=210-21=189\). Since the diagonals from 1 particulat vertex shouldn't be counted then \(189-(21-3)=171\) (one vertex makes \(n-3\) diagonals).

Re: How many diagonals does a polygon with 21 sides have... [#permalink]

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23 Sep 2010, 04:04

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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 diagnols

Alternate Solution

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 20-sided 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 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 = \((21*20/2)-21=210-21=189\)

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: \(189-18=171\) 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 21-3=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.
_________________

Re: How many diagonals does a polygon with 21 sides have... [#permalink]

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23 Sep 2010, 06:47

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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 20-sided 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 !!
_________________

Polygon problems a sucking combination---HELP! [#permalink]

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10 Nov 2010, 06:59

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?

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 10401 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
_________________

Re: How many diagonals does a polygon with 21 sides have, if one [#permalink]

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12 Oct 2013, 05:22

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Re: How many diagonals does a polygon with 21 sides have... [#permalink]

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10 Dec 2013, 20:46

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 \(C^2_n-n=\frac{n(n-3)}{2}\): \(C^2_n\) 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 \(C^2_n-n=210-21=189\). Since the diagonals from 1 particulat vertex shouldn't be counted then \(189-(21-3)=171\) (one vertex makes \(n-3\) 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

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
_________________

Re: How many diagonals does a polygon with 21 sides have... [#permalink]

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10 Dec 2013, 22:11

VeritasPrepKarishma wrote:

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

Wow, it's nice explanation. Your way is much more understandable. Thanks.

Re: How many diagonals does a polygon with 21 sides have, if one [#permalink]

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26 Jun 2014, 03:22

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 \(C^2_n-n=\frac{n(n-3)}{2}\): \(C^2_n\) 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 \(C^2_n-n=210-21=189\). Since the diagonals from 1 particulat vertex shouldn't be counted then \(189-(21-3)=171\) (one vertex makes \(n-3\) 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 (18-3) 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 (n-3)/2 & Each vertex sends of n-3 diagonals #desired diagonals = #diagonals with 18 sides - #diagonals 3 vertices would send = 18 (18-3) / 2 - (18-3) * 3 = 135 - 45 = 90 ---------------------------------------> (2)

Could you please tell why I see this difference in the answers (1) and (2)?

Q: How many diagonals does a polygon with 18 sides have if three of its vertices do not send any diagonal? A: #lines with (18-3) 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 (n-3)/2 & Each vertex sends of n-3 diagonals #desired diagonals = #diagonals with 18 sides - #diagonals 3 vertices would send = 18 (18-3) / 2 - (18-3) * 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.
_________________

Re: How many diagonals does a polygon with 21 sides have, if one [#permalink]

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04 Jul 2014, 22:52

VeritasPrepKarishma wrote:

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 (18-3) 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 (n-3)/2 & Each vertex sends of n-3 diagonals #desired diagonals = #diagonals with 18 sides - #diagonals 3 vertices would send = 18 (18-3) / 2 - (18-3) * 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.

Re: How many diagonals does a polygon with 21 sides have, if one [#permalink]

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26 Aug 2014, 12:00

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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

gmatclubot

Re: How many diagonals does a polygon with 21 sides have, if one
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26 Aug 2014, 12:00

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