# Problem on permutations and combinations | Set 2

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^{m}C_{3}= 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 = 1Input : 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 = 2Input: m = 5

Output: 5

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

Answer : (

^{m}C_{2}*^{n}C_{2}).

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

=^{2}C_{2}*^{2}C_{2}

= 1 * 1 = 1Input: m = 4, n = 4

Output: 36

**There are â€˜nâ€™ points in a plane, out of which â€˜mâ€™ points are co-linear. Find the number of triangles formed by the points as vertices ?**

Number of triangles =

^{n}C_{3}–^{m}C_{3}

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^{n}C_{3}ways.

Number of triangles formed by 10 points when no 3 of them are co-linear =^{10}C_{3}……(i)

Similarly, the number of triangles formed by 4 points when no 3 of them are co-linear =^{4}C_{3}……..(ii)Since triangle formed by these 4 points are not valid, required number of triangles formed =

^{10}C_{3}–^{4}C_{3}= 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 :

^{n}C_{2}–^{m}C_{2}+ 1

Explanation : Number of straight lines formed by n points when none of them are col-linear =^{n}C_{2}

Similarly, the number of straight lines formed by m points when none of them collinear =^{m}C_{2}

m points are collinear and reduce in one line. Therefore we subtract^{m}C_{2}and add 1.

Hence answer =^{n}C_{2}–^{m}C_{2}+1

Examples:

Input : n = 4, m = 3

output : 1

We apply this formula

Answer =^{4}C_{2}–^{3}C_{2}+1

= 3 – 3 + 1

= 1

**More practice questions on permutation and combination :**

Quiz on Permutation and Combination

Combination and Permutation Practice Questions

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