pnc doubt
A regular polygon of 10 sides is constructed. No. of ways 3 vertices be selected so that no two vertices are consecutive is.....
i understand this and but i tried using a diff method and it didnt come right...if i arrange the 7 gaps between 3 vertices ensuring theres always one in between 2 vertices i'll get x + y + z =4 and so i'll get 6c2 = 15...why isnt this approach correct?
16 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.sorry for the double ping 🙏
i dont think that will be possible
what abotu when 2 sides are consecutive it doesnt count that
i mean 2 points are consecutive
and there are also those triangles for which all 3 points are consecutive
at corners
no 2 vertices should be consecutive right
so i just arranged the 7 vertices
between any 3
oh wait
😭
i forgot to choose those 3 first
they can
its in the question
nvm am dumb
no worries 😭 im dumb too
hmmm
accha i think i got it
10 ways to select a starting point...then 15 ways for division of gaps...divided by 3 since vertices arent distict
okie so from what i could get, you're trying to place 3 vertices with at least one gap between them but that doesn't account for all positions correctly. Instead think of it like thi....place the 7 empty spots first and then pick 3 out of the remaining 7 positions for the vertices.
hm?
all g?