Permutation and Combination Notes for SSC Exams
Rahul Kumar
SSC Exam Expert & Content Editor
In this note
- 1.Factorial
- 2.Fundamental Principles of Counting
- 3.Permutation and Combination: Formulas
- 4.Permutation with Repetition and Alike Items
- 5.Solved Examples
- 6.Common Formulas for Quick Use
- 7.Quick Tips
- 8.Quick Revision Points
- 9.Practice Questions With Answers
- 10.How to Practise This Chapter
- 11.Frequently Asked Questions
PDF Notes: Permutation and Combination Notes for SSC Exams
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Permutation and combination teach how to count the number of ways of arranging or selecting things. Permutation is used when the order matters, and combination is used when the order does not matter. These permutation and combination notes for SSC explain the formulas, the difference between the two and the common question types with solved examples and practice questions in simple language.
Factorial
- n! = n × (n − 1) × (n − 2) × ... × 2 × 1.
- 0! = 1 and 1! = 1.
- 5! = 120, 6! = 720 and 7! = 5040.
Fundamental Principles of Counting
| Principle | Meaning | Example |
|---|---|---|
| Multiplication rule | If one work can be done in m ways and another in n ways, both together can be done in m × n ways | 3 shirts and 4 pants give 12 dresses |
| Addition rule | If one work can be done in m ways or another in n ways (not both), the total is m + n | Going by 3 buses or 2 trains gives 5 ways |
Permutation and Combination: Formulas
| Term | Formula | Use |
|---|---|---|
| Permutation | nPr = n! / (n − r)! | Arrangement; order matters |
| Combination | nCr = n! / [r! (n − r)!] | Selection; order does not matter |
| Relation | nPr = nCr × r! | Link between the two |
| Symmetry | nCr = nC(n − r) | Selecting r is the same as leaving n − r |
| Special values | nC0 = nCn = 1 and nC1 = n |
Permutation with Repetition and Alike Items
| Case | Formula |
|---|---|
| Arrangement of n items of which p are alike of one kind and q are alike of another | n! / (p! q!) |
| Arrangement of n different items in a line | n! |
| Arrangement of n items in a circle | (n − 1)! |
| Necklace or garland (clockwise and anticlockwise same) | (n − 1)! / 2 |
| n items taken r at a time with repetition allowed | nr |
Solved Examples
Example 1. Find the value of 6P3.
6P3 = 6 × 5 × 4 = 120.
Example 2. Find the value of 5C2.
5C2 = (5 × 4) / (2 × 1) = 10.
Example 3. In how many ways can the letters of the word BANANA be arranged?
There are 6 letters, with A repeated 3 times and N repeated 2 times. Ways = 6! / (3! × 2!) = 720 / 12 = 60.
Example 4. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 without repetition?
5P3 = 5 × 4 × 3 = 60.
Example 5. A committee of 3 is to be chosen from 5 men and 4 women, with at least one woman. In how many ways can it be done?
Total ways = 9C3 = 84. Ways with no woman = 5C3 = 10. So the answer = 84 − 10 = 74.
Example 6. In how many ways can 5 persons sit around a round table?
(5 − 1)! = 4! = 24.
Common Formulas for Quick Use
| Situation | Formula |
|---|---|
| Handshakes among n persons (each pair once) | nC2 = n(n − 1) / 2 |
| Diagonals of a polygon with n sides | nC2 − n = n(n − 3) / 2 |
| Number of triangles from n points (no three in a line) | nC3 |
| Number of lines from n points (no three in a line) | nC2 |
Quick Tips
- Ask "does the order matter?" first: yes means permutation, no means combination.
- For "at least" questions, subtract the opposite case from the total.
- For words with repeated letters, divide by the factorial of the repeated count.
- Remember that arrangements around a circle use (n − 1)!, not n!.
Quick Revision Points
- nPr = n! / (n − r)! and nCr = n! / [r! (n − r)!].
- nPr = nCr × r!.
- nCr = nC(n − r).
- Circular arrangement of n items = (n − 1)!.
- Handshakes among n persons = n(n − 1) / 2.
Practice Questions With Answers
- Find the value of 7C3.
Answer: 35 - Find the value of 5P2.
Answer: 20 - In how many ways can 4 persons be arranged in a row?
Answer: 24 - How many arrangements of the letters of the word LEVEL are possible?
Answer: 30 (5! / (2! × 2!)) - In how many ways can 6 persons sit around a round table?
Answer: 120 - How many handshakes are possible among 10 persons?
Answer: 45 - How many diagonals does a hexagon have?
Answer: 9 - A team of 2 is to be chosen from 6 players. In how many ways can it be done?
Answer: 15 - How many 2-digit numbers can be made from 1, 2, 3, 4 without repetition?
Answer: 12 - Find the value of 8C6.
Answer: 28 (same as 8C2)
How to Practise This Chapter
Solve five problems daily from different types: value of nPr and nCr, words, committees and circular arrangements. Then attempt timed sets from our practice papers and the previous year papers page. Read the previous chapter, Probability, and see all topics on the Quantitative Aptitude notes page. For arrangement-based reasoning, also read the Reasoning notes.
Frequently Asked Questions
What is permutation?
Permutation is the number of ways of arranging items in which the order of the items matters.
What is combination?
Combination is the number of ways of selecting items in which the order does not matter.
What is the formula of nPr?
nPr equals n factorial divided by (n minus r) factorial.
What is the formula of nCr?
nCr equals n factorial divided by r factorial times (n minus r) factorial.
What is the difference between permutation and combination?
In permutation, the order matters, as in arranging a word. In combination, the order does not matter, as in choosing a team.
What is the value of 0 factorial?
The value of 0! is 1.
How do I arrange letters of a word with repeated letters?
Divide the factorial of the total letters by the factorial of the number of times each letter repeats.
What is circular permutation?
It is the arrangement of items around a circle. For n items, the number of ways is (n minus 1) factorial.
How do I solve "at least one" selection problems?
Find the total number of ways and subtract the number of ways in which none of that kind is selected.
How many handshakes are there among n persons?
The number of handshakes is nC2, which equals n (n minus 1) divided by 2.
Is permutation and combination difficult?
No. It becomes easy if you decide whether the order matters and then apply the right formula.
How many questions come from permutation and combination in SSC exams?
Usually one or two questions come from this chapter in SSC Quantitative Aptitude papers, but the number changes from year to year.
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Probability Notes for SSC Exams
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Chain Rule and Direct and Inverse Variation: Formulas, Tricks and Solved Examples for SSC CGL, CHSL and MTS
Rahul Kumar
SSC Exam Expert & Content Editor
Rahul is a senior SSC exam strategist with 8+ years of experience helping aspirants crack CGL, CHSL and MTS exams. He writes in-depth notification breakdowns, exam pattern guides and preparation strategies.