Skip to content
Related Articles

Related Articles

Improve Article

Problem on permutations and combinations | Set 2

  • Difficulty Level : Easy
  • Last Updated : 25 May, 2018

Prerequisite : Permutation and Combination

Given a polygon of m sides, count number of triangles that can be formed using vertices of polygon.

Answer : [m (m – 1)(m – 2) / 6]
Explanation : There are m vertices in a polygon with m sides. We need to count different combinations of three points chosen from m. So answer is mC3 = m * (m – 1) * (m – 2) / 6
Examples :
Input: m = 3
Output: 1
We put value of m = 3, we get required no. of triangles = 3 * 2 * 1 / 6 = 1

Input : m = 6
output : 20

Given a polygon of m sides, count number of diagonals that can be formed using vertices of polygon.



Answer : [m (m – 3)] / 2.
Explanation : We need to choose two vertices from polygon. We can choose first vertex m ways. We can choose second vertex in m-3 ways (Note that we can not choose adjacent two vertices to form a diagonal). So total number is m * (m – 3). This is twice the total number of combinations as we consider an diagonal edge u-v twice (u-v and v-u)
Examples:
Input m = 4
output: 2
We put the value of m = 4, we get the number of required diagonals = 4 * (4 – 3) / 2 = 2

Input: m = 5
Output: 5

Count the total number rectangles that can be formed using m vertical lines and n horizontal lines

Answer : (mC2*nC2).
We need to choose two vertical lines and two horizontal lines. Since the vertical and horizontal lines are chosen independently, we multiply the result.
Examples:
Input: m = 2, n = 2
Output: 1
We have the total no of rectangles
= 2C2*2C2
= 1 * 1 = 1

Input: m = 4, n = 4
Output: 36

There are ‘n’ points in a plane, out of which ‘m’ points are co-linear. Find the number of triangles formed by the points as vertices ?

Number of triangles = nC3mC3
Explanation : Consider the example n = 10, m = 4. There are 10 points, out of which 4 collinear. A triangle will be formed by any three of these ten points. Thus forming a triangle amounts to selecting any three of the 10 points. Three points can be selected out of the 10 points in nC3 ways.
Number of triangles formed by 10 points when no 3 of them are co-linear = 10C3……(i)
Similarly, the number of triangles formed by 4 points when no 3 of them are co-linear = 4C3……..(ii)

Since triangle formed by these 4 points are not valid, required number of triangles formed = 10C34C3 = 120 – 4 = 116

There are ‘n’ points in a plane out of which ‘m’ points are collinear, count the number of distinct straight lines formed by joining two points.

Answer : nC2mC2 + 1
Explanation : Number of straight lines formed by n points when none of them are col-linear = nC2
Similarly, the number of straight lines formed by m points when none of them collinear = mC2
m points are collinear and reduce in one line. Therefore we subtract mC2 and add 1.
Hence answer = nC2mC2 +1
Examples:
Input : n = 4, m = 3
output : 1
We apply this formula
Answer = 4C23C2 +1
= 3 – 3 + 1
= 1

More practice questions on permutation and combination :
Quiz on Permutation and Combination
Combination and Permutation Practice Questions

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :