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# Permutation and Combination Fundamentals

#### Learn the fundamental principles of Permutation and Combination. This article will equip you to solve problems on this topic appearing across all competitive examinations in a time bound manner.

In this article, we will discuss the basic concepts of Permutation & Combination and formulas required for solving problems on the same. This topic is covered in all competitive exams, so you cannot afford to take the risk of avoiding it. After reading this article, you will become familiar with the basics concepts of permutation and combination.

Permutation and Combination formula

• Permutation Formula
Permutation is defined as arrangement of r things that can be done out of total n things. This is denoted by nPr which is equal to n!/(n-r)!
• Combination formula
1. Combination is defined as selection of r things that can be done out of total n things. This is denoted by nCr which is equal to n!/r!(n-r)!
2. As per the Fundamental Principle of Counting, if a particular thing can be done in m ways and another thing can be done in n ways, then either one of the two can be done in m + n ways and both of them can be done in m × n ways.

Solved Permutation and Combination Problems

Example 1: How many four-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 (Repetition of digits not allowed)?

Solution: Thousand's place can be filled in 6 ways. Hundred's place can be filled in 5 ways. Ten's place can be filled in 4 ways. Unit's place can be filled in 3 ways. So, using the Fundamental Principle of Counting, we get the answer as 6 × 5 × 4 × 3 = 360. Or using formula of Permutations, we need to arrange 4 digits out of total 6 digits. This can be done in 6P4 = 360 ways.

Example 2: A person has 6 friends to be invited for dinner through invitation cards, and he has 3 servants. In how many ways can he extend the invitation card?

Solution: We can see that the 1st friend has 3 options to receive the card, i.e. either from 1st servant or 2nd or 3rd. Similarly 2nd friend also has 3 options to receive the card, i.e. either from 1st servant or 2nd or 3rd. So we can say that each of the 6 friends has 3 options to receive the card. Hence the answer would be 3 × 3 × 3 × 3 × 3 × 3 = 36 = 729 ways.

Example 3: There are 10 questions in an exam. In how many ways can a person attempt at least one question?

Solution: A person can attempt 1 question or 2 questions or .....till all 10 questions. One question out of ten questions can be attempted in 10C1 = 10 ways. Similarly two questions out of ten questions can be attempted in 10C2 = 45 ways. Going ahead by the same logic, all ten questions can be attempted in 10C10 = 1 way. Hence the total number of ways = 10 + 45 + 120 +.....10 + 1 = 1023 ways (Using the formula of Combination).

Alternate Method:

Or some logic can be applied: Every question has 2 options, either it is attempted or not. Going ahead with this logic, since there are 10 questions, and each question has 2 options, so total number of cases = 210= 1024. But this count includes one case in which no question is attempted. This is the violation of the information given. So this case needs to be subtracted. Hence the total number of cases would be 1024 - 1 = 1023.

Master the concepts of P & C by watching this P & C Achievers video on the fundamentals of the topic

Permutation and Combination: Key Learnings

• In this article, you learned the basic concepts and formulae useful for solving questions on Permutations & Combinations.
• Statement questions are primarily designed to test your understanding to frame an equation and then use formulae on it.

For any query on any concept or example mentioned, ask us! Post in the section below.

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