NDA Maths Permutations and Combinations: Important Questions, Tricks & PYQs (2026)

Permutations and Combinations form one of the most important and scoring chapters in NDA Mathematics. Questions from this topic are frequently asked in exams and are often based on direct formulas and logical application. With the right approach, students can solve even complex problems in seconds. Check important concepts, formulas, tricks, and PYQs-style questions to help you build strong command over the topic and improve your exam performance.

Number of Lines from N Points

The number of lines that can be drawn using n points in a plane is a key concept.

Question: To find the number of straight lines that can be drawn from 10 points, out of which 7 are collinear, the calculation is 10C2 - 7C2 + 1.

Number of Points of Intersection Between Non-Concurrent, Non-Parallel Lines

The number of points of intersection between n non-concurrent, non-parallel lines is given by nC2.

Question: For the maximum number of points of intersection of eight straight lines, the formula is 8C2.

Number of Triangles from N Points

To find how many triangles can be formed using n points, consider the following cases:

Question 1 (Collinear Points): The number of triangles that can be formed from 12 points, out of which 7 are collinear, is 12C3 - 7C3.

Question 2 (Points on Sides of a Triangle): Consider a scenario where points are on the sides of a triangle: 3 points on one side, 4 points on another side, and 5 points on the third side. The total number of points is 3 + 4 + 5 = 12. The number of triangles that can be constructed using these points as vertices is calculated as 12C3 - (3C3 + 4C3 + 5C3). (Here, 3C3, 4C3, and 5C3 represent combinations of collinear points on each side that cannot form a triangle).

Number of Quadrilaterals from N Points with Collinear Points

For the number of quadrilaterals, if there are n points and r points are collinear, the RR method formula is:

nC4 - rC3 * n * (n-r)C1 - rC4. This formula might seem complicated to remember, but it is important to memorize.

Number of Parallelograms from Intersecting Parallel Lines

The number of parallelograms formed if m parallel lines intersect n parallel lines can be found using these methods:

1. mC2 * nC2

2. (m * (m-1) * n * (n-1)) / 4

Question: If there are 5 parallel straight lines (m=5) and 4 parallel straight lines (n=4) intersecting, using the second method:

(5 * 4 * (5-1) * (4-1)) / 4 = (5 * 4 * 4 * 3) / 4 = 60. These types of questions are often Previous Year Questions (PYQs).

Number of Triangles in a Polygon with No Common Sides

The number of triangles in a polygon of n sides that have no common sides with the polygon is given by the formula:

(n * (n-4) * (n-5)) / 6.

Question: For a regular polygon with 10 sides, the number of triangles with no common points (sides) is:

(10 * (10-4) * (10-5)) / 6 = (10 * 6 * 5) / 6 = 50.

Maximum Number of Points of Intersection (Circles)

If n circles intersect, the maximum number of points of intersection will be 2 * nC2.

Question 1: The maximum number of points of intersection of eight circles is 8C2 * 2.

Question 2: The maximum number of points of intersection of 10 circles is 10C2 * 2. This is a PYQ.

Maximum Number of Points of Intersection (M Straight Lines and N Circles)

The greatest number of points of intersection made by m straight lines and n circles is a result that students must remember. Questions have been asked on this topic.

Question: A problem might ask for the maximum number of points of intersection of four circles and four straight lines. The RR method for this type of problem is crucial for competitive exams.

Number of Diagonals in a Polygon

The number of diagonals for a polygon with n sides is given by the formula:

n * (n-3) / 2.

Question: If a polygon has 44 diagonals, we need to find the number of its sides.

n * (n-3) / 2 = 44

n * (n-3) = 88

By checking common values, if n=11, then 11 * (11-3) = 11 * 8 = 88. Thus, the number of sides n = 11.

Sum of All Numbers Formed by Given Digits

To find the sum of all numbers formed by given digits, use the formula:

S * (n-1)! * R

Where:

• S represents the sum of the digits.

• n represents the number of digits.

• R represents a number consisting of n ones (e.g., if n=4, R=1111).

Question: Find the sum of all possible numbers greater than 2000 formed using the digits 2, 3, 4, 5. (All numbers formed from these digits will be greater than 2000).

• Number of digits n = 4.

• Sum of digits S = 2 + 3 + 4 + 5 = 14.

• Using the formula: 14 * (4-1)! * 1111 = 14 * 3! * 1111 = 14 * 6 * 1111.

Sum of All Numbers Formed by Digits (Including Zero)

When zero is included in the set of given digits, the calculation involves two steps:

1. Total Combinations: Calculate the sum using the formula S * (n-1)! * R as if zero were a regular digit (using the total n for number of digits).

2. Subtract Invalid Combinations: Subtract the sum calculated for cases where zero would be the leading digit. This is done by effectively considering a set of n-1 digits (excluding zero) for a sub-calculation. This method is highlighted as one of the best RR methods to save time in exams.

Question: If the digits are 0, 1, 4, 5:

• Sum of digits S = 0 + 1 + 4 + 5 = 10.

• The calculation involves two parts: first using n=4 (considering all digits including zero), and then subtracting a calculation using n=3 (effectively removing permutations where zero leads).

Total Number of Games/Matches

For questions related to games or matches, the total number of matches can be found using the formula:

(1/2) * n * (n-1) * p

Where:

• n represents the number of teams.

• p represents the number of times one team plays with another team.

Question: If teams play one match with each other (so p=1), and the total number of games is 36, we need to find 'n'.

36 = (1/2) * n * (n-1) * 1

72 = n * (n-1). This equation can be solved to find the value of 'n'.

Tricks and Shortcut Methods (RR Method)

This is the most important section for exam preparation.

1. Lines from Points Trick

Normal case: nC2
With collinear points: nC2 – rC2 + 1

2. Triangle Formation Trick

Total triangles: nC3
With collinear points: nC3 – rC3

3. Points on Sides Trick

Total combinations – invalid collinear combinations

4. Parallel Lines Trick

Number of parallelograms = mC2 × nC2
Alternate formula: (m × (m-1) × n × (n-1)) / 4

5. Polygon Tricks

Diagonals = n(n-3)/2
Triangles with no common side = n(n-4)(n-5)/6

6. Circles Intersection Trick

Maximum points = 2 × nC2

7. Sum of Numbers Trick

Formula: S × (n-1)! × R

8. Zero Digit Trick

Solve normally, then subtract cases where 0 comes first

9. Matches Trick

Total matches = (n(n-1)/2) × p