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# How many diagonals does a polygon with 21 sides have, if one of its ve

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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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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
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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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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 = $$(21*20/2)-21=210-21=189$$

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.
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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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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).

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
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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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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
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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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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).

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.
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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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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.
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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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formula for counting diagonal in a polygon n(n-3)/2
s ; 21*18/2 ; 189
since 1 diagonal does not connect so for 1 diagonal 21-3 ; 18
formula for diagonal from 1 point ( n-3)
189-18 ; 171
IMO C

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. 171
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 = $$(21*20/2)-21=210-21=189$$

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?
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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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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. 171
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 = $$(21*20/2)-21=210-21=189$$

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?

Asked: How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?

The number of such diagonals = 20C2 - (21-2) = 20*19/2 - 19 = 190 - 19 = 171

IMO C
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How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
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First, consider the left and right points that the leftover point has. They don't bother about the restrictions. They will form 21-3 = 18 diagonals each, i.e., 18*2 = 36 in total (We'll take care of repetition later on). Now, points that are not adjacent to the leftover point will form 17 diagonals each. As we have 18 points left, this time we'll get 18*17 = 306 diagonals. Total = 306 + 36 = 342. As each diagonal is counted twice, the final answer is = 342/2 = 171.
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Re: How many diagonals does a polygon with 21 sides have, if one of its ve [#permalink]
KarishmaB 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.

If I said that regarding this exemple and if we continue with the formula 18C2 and at the end we conclude by 135 - 3*(18-3) - 1
Is it correct ?
What will happens if we decide to try with 4 vertices removed ?
Am I reason if I said that it is : 18C2 - 18 -[4*(18-3)-1] ?

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