1Doing it in another method, can the answer possibly be 5040 ??
Quadrilaterals are formed by using the vertices of a convex polygon of 24 sides. The number of quadrilaterals having atleast one side of the quadrilateral in common with the side of the polygon is ?
We can find out the total no. of quadrilaterals that can be formed and subtract from it the no. of quadrilaterals with no side common to that of the polygon.
24C4 - 12C4 = 10131
I got 12 by choosing alternate vertices.
I don't get the correct answer this way :|
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28 Answers
1@ Kunl : Tum thoda aur help kar do toh dimaag chalna shuru karega ! [115] [4]
1@ AKHIL : Oh ha I think I got what u want to say.....i applied that in the other two cases as well , and the answer is coming out to be
(24x20C2) + (24x 19C1) + (24x1) = 5040... (as i had asked in a previous reply).. [317]
6no no i m not talkin abt the left ones...........
@kunl
i still think its 24!!!:P
1@ AKHIL :
I purposely left the vertices adjacent to the selected side , because i am considering quadrilaterals with EXACTLY one side common.
By joining one of the other 20 vertices with one of the selected vertices , we do not obtain a side of the polygon , so what's the use excluding them ?
[308]
6@abhishek
when u r finding the no. of quadrilaterals with one side common , u did 20C2 which i think is perfectly fine...........
but the 2 vertices u left can also be chosen in 24 ways!!!
think abt that!!
21we will use B.P
Number of quadrilaterals with no side common = Number of ways of choosing 4 among 24 letters kept in a circle so that no two are r consecutive = Number of ways of choosing 4 numbers from the set {1,2,....23} so that no two are consecutive + Number of ways of choosing 3 numbers from {1,2,..21}so that no two are consecutive
Done !
49let me try my hand at this..
Considering a polygon with a centre of symmetry!
The selection being circular in nature,
All possible quadrilateral is 23C3
now number of quadrilaterals having no common side with polygon..
since, each vertex has equivalent position, first vertex can be chosen in only one way.
Having chosen one vertex, we can name other vertices as 1,2,3,4,5....23, i.e. differentiate each from the other.
let the second vertex was vertex number x, third was vertex number y, fourth was vertex number z.
x≥2
y≥x+2
z≥y+2
23≥z+1
converting inequality to equality,
x=2+a
y=x+2+b
z=y+2+c
23=z+1+d
a,b,c,d all are members of the family of whole numbers!
adding up the equalities,
23=7+a+b+c+d
gives, 16=a+b+c+d
no of solutions of form (x,y,z)=number of solutions of form (a,b,c,d)=19C3
SO REQD. NO OF QUADRILATERALS= 23C3 - 19C3
1dumb ricky
my answerr is coming as 419...and i am also very sure of it!
1The answer is somewhat less than 210 , that I am sure of . But the exact value still eludes me .
EDIT - I think the answer is 191 .
1Abhishek....I found that you got very close to the answer..think more.
1**For quadrilaterals with exactly one side common:
We fix up one side i.e. two vertices.
Now we exclude the sides adjacent to the fixed side (i.e we remove 2 vertices) , and then select two vertices out of the remaining 20 vertices, in 20C2 ways.
**For quadrilaterals with exactly two sides common:
We fix up two side i.e. three vertices.
Now we exclude the sides adjacent to the fixed side (i.e we remove 2 vertices) , and then select one vertex out of the remaining 19 vertices, in 19C1 ways.
**For quadrilaterals with exactly three sides common:
I think that after fixing up 3 sides, the 4th side is automatically generated...so one way.
So total number of ways = 20C2 + 19C1 + 1 = 210.
BUT I HAVE DOUBTS ABOUT THIS ANSWER...HELP ! [12]