# Permutation and Combination in Geometry

It is quite difficult to quantify the importance of P and C in geometry, a good number of P and C questions which use the concepts of geometry (and vice-versa) have been asked in the CAT and other premier B-school exams.

Example-1

How many diagonals will be there in an

*n*-sided regular polygon?Solution

An

*n*-sided regular polygon will have n vertices. And when we join any of these two vertices (*C*^{n}_{2}) we get a straight line, which will be either a side or a diagonal.So,

*C*^{n}_{2}= Number of sides + number of diagonals=

*n*+ number of diagonalsHence, the number of diagonals =

*C*^{n}_{2 }âˆ’*n*=

*Above written result can be used as a formula also.*Example-2

Ten points are marked on a straight line and 11 points are marked on another parallel straight line. How many triangles can be constructed with vertices among the above points?

Solution

Triangles will be constructed by taking one point from the 1st straight line and two more points from the 2nd straight line, and vice versa.

So, the total number of âˆ† formed =

^{10}C_{2}Ã—^{11}C_{1}+^{11}C_{2}Ã—^{10}C_{1}= 1045Example-3

There is an

*n*-sided polygon (*n*>5). Triangles are formed by joining the vertices of the polygon. How many triangles can be constructed which will have no side common with the polygon?Solution

An

*n*-sided polygon will have*n*vertices. Triangles constructed out of these*n*vertices will be of three types:- Having two sides common with the polygon â€” Out of total
*n*vertices, any combination of three consecutive vertices will give us the triangle which is having two sides common with polygon =*n* - Having one side common with the polygon â€” Number of selection of three vertices out of which two are consecutive (If we select A
_{5}and A_{6}as the two vertices, then A_{7}or A_{4}should not be the third vertex because it will constitute the two sides of the common triangle). =*n*Ã—^{(nâˆ’4)}C_{1} - Having no side common with the polygon. And total number of triangles formed will be
C^{n}_{3}.C^{n}_{3}âˆ’*n*Ã—^{(nâˆ’4 )}C_{1}â€“*n*

**Some More Important Results:**

- Maximum No.of points of Intersection among
*n*Straight Lines =C^{n}_{2} - Maximum No. of points of Intersection among
*n*Circles =P^{n}_{2}