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.

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.

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

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.

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,

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.

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.

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

Why is the title "Probability"? Probably to confuse others...

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