I have just learnt permutations, dispositions, combinations. How deserve to I fix it with these concepts? I drew it and it was $35$ diagonals. How can I prove it with this method?



A diagonal join a peak to one of the vertices that perform not incorporate that crest itself and also the immediately adjacent vertices. So: for each vertex there are 7 diagonals. Times 10 amounts to 70; each diagonal is counted twice, so the final answer is 35.

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Now, utilizing combinations and such: There space $inom102;$("10 select 2") pairs of vertices, which amounts to 45. For this reason there room 45 line segments joining bag of vertices. Precisely 10 that those room sides of the decagon, the others are diagonals. Answer: 35. (Corrected; original had "10 choose 9" because that no reason various other than my lack of concentration.)


Formula because that calculating variety of diagonals of any type of polygon of n sides = n*(n - 3)/2

So here it"s a decagon ,that is a 10 face polygon, for this reason n = 10. Simply plug value of n right into the formula , girlfriend get: 10*(10-3)/2 = 35. Ans :)

(Note : no matter what face polygon the is, you deserve to find any kind of no the diagonals in any type of polygon)


(N-1 select 2 ) -1

Example, (10-1 pick 2) -1= (9 select 2) -1= 36-1= 35 diagonal line lines

This will occupational for any type of regular shape

Example 2/ 20 sides would be C19,2 -1=170 Check: C20,2 - 20 =170

Very straightforward formula emerged by Shawn Covrigaru


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