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Probability

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by newton9 » Sun Mar 28, 2010 10:57 am
32. 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


My approach is:

Total straight lines connecting 21 points is 21C2 = 210.

Of these 21 are actual sides of polygon. So number of diagonals is 210-21 = 189.

Since one of the vertices is not connected, total number of diagonals missing = 18 ( substracting the two adjacent points and the point itself).

So I ended up at 171. But the answer is 170. Did I miss anything?

Any other better approaches??
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by srinivasarajui » Mon Mar 29, 2010 4:08 am
newton9 wrote: Since one of the vertices is not connected, total number of diagonals missing = 18 ( substracting the two adjacent points and the point itself).
There are 21 vertices, So for each point there are 19 diagonals(21-2). So now you get the correct answer.
Srinu
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by pops » Mon Mar 29, 2010 6:38 am
srinivasarajui wrote:
newton9 wrote: Since one of the vertices is not connected, total number of diagonals missing = 18 ( substracting the two adjacent points and the point itself).
There are 21 vertices, So for each point there are 19 diagonals(21-2). So now you get the correct answer.
Hey, I dont agree. Can you please check:

number of vertices = 21
lines from each vertex = 20
out of these 20, 2 are to adjacent vertices i.e. 2 sides and these are not diagonals
hence, number of diagonals from each vertex is 18

so 21C2 - 21 - 18 = 171 is indeed correct.
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by eaakbari » Mon Mar 29, 2010 1:31 pm
Each vertex in any polygon has an option of forming a diagonal with any vertex apart from its neighbouring two

Hence we can express each vertex to have n -3 options (1 for itself and its 2 neighbours)
But number of diagonals in polygon will be n(n-3)/2 as the same diagonal will be counted from 2 vertexes

Now for this 21 sided there are 21 vertexes, since 1 is isolated
we have 21 vertexes
Putting in above formula( self derivation m a little proud)
Its 20*17/2

which gives 170
Hence B
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by akhpad » Tue Mar 30, 2010 2:08 am
Here, I am also agree with 171.

eaakbari mistakenly got 170 because he calculated for only 20 vertices.

Here, 21 vertices but one in not participating.
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by eaakbari » Tue Mar 30, 2010 5:22 am
The deal is if one vertex is not participating, we cannot get any diagonal from there. So it is as good as 20 vertices. Do go through my explanation, I do believe its right.

@akhp77, It was not a mistake mate,
:wink:
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by kstv » Tue Mar 30, 2010 6:42 am
newton9 wrote:32. 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
My approach is:
Total straight lines connecting 21 points is 21C2 = 210.
Of these 21 are actual sides of polygon. So number of diagonals is 210-21 = 189.
Since one of the vertices is not connected, total number of diagonals missing = 18 ( substracting the two adjacent points and the point itself).
So I ended up at 171. But the answer is 170. Did I miss anything?
Any other better approaches??
Such a great approach. Just tried to work it out with a Pentagon.
5c2 = 10 no of st. lines. 5 are the sides of the polygon. If you omit the lines from one of the vertices there will be 2 diagonals less. Drawing a Pentagon is easier and eassier to make out that from one point you can connect four other points but two of them are sides of the pentagon. Ans - three diagonals (10-5-2) can be drawn.
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by akhpad » Tue Mar 30, 2010 7:43 am
Let me explain to eaakbari.

No of diagonals = n(n-3)/2, where n is the no of vertices. This formula is correct and it can be derived from the same approach what newton9 has mentioned.
Lets take a pentagon, which has five diagonals [ 5*(5-3)/2 = 5 ].
If one vertex is not participating, then no of diagonals = 4*(4-3)/2 = 2 from your approach. See here, the no of diagonals is 2, which is same as a quadrilateral have. Because we considered no of vertices is 4.

But in pentagon, if one vertex is not participating then it has 3 diagonals. Please take a closer look about your mistake.
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by eaakbari » Wed Mar 31, 2010 3:59 am
Oh yes. Thank you akhp77, i do get it now. Let me correct my approach.

No of diagonals for 21 sided figure = n (n-3)/2
= 21(18)/2
= 189

Now since one vertex is not participating, the number of diagonals it could have made is n-3 (excluding itself and two neighbour vertices)
Hence 19

so now 189 - 19 = 170
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by sanju09 » Fri Apr 02, 2010 12:54 am
Why is the title "Probability"? Probably to confuse others...

https://www.beatthegmat.com/how-many-dia ... 45438.html
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