Permutation and Combination

Number of Permutations

Permutations are a very important topic of most exams. In every exam that is based on reasoning ability, you shall find a few questions about Permutations. In mathematics, permutations are a way of counting the number of arrangements. Here we will see what we mean by counting arrangements and the formula for getting the number of Permutations. Let us see!

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To understand the formula for permutations we will first see how we count the arrangements. Let us state what we call the fundamental principle of counting. Suppose an event E, can occur in ‘m’ different ways. Let us say something like we are arranging a ball in 12 empty slots. So ‘m’ here will be 12. Next, suppose corresponding to each instance of the event E, another event F can occur in ‘n’ different ways. Then the total number of ways in which the two events can occur is m×n.

For example, in the case of the ball that is to be arranged in 12 empty slots, let us say we have another ball. For this ball, we will have only 11 empty slots left. So the number of different ways that we can arrange this ball in is 11. One empty slot corresponding to each arrangement. What is the number of ways in which two balls can be arranged into 12 empty slots? The answer is 12×11 in accordance with the fundamental principle of counting.

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Some Examples On The Counting Principle

Example 1: Salman has 5 shirts, 6 shoes and 4 pants from which he has to choose an outfit. In how many ways can he select one top, one skirt and one cap?

Answer: The fundamental principle of counting states that if Salman can wear his shoes in 6 ways, and his shirts in 5 ways, and his pants in 4 ways, then the total number of ways in which he can select a different outfit will be given by the product 6×5×4 = 120. Therefore the total number of different selections or dresses that can be chosen are 120. These choices are what we call the Permutations.

Example 2: How many multiples of 5 are there from 10 to 95?
Answer: The multiples of 5 are integers having 0 or 5 in the digit in the unit’s place. The first digit from the right can be chosen in 2 ways. The second digit can be any one of 1,2,3,4,5,6,7,8,9 digits. Thus there are a total of 9 ways to choose the second digit. Thus, there are 2×9 =18 multiples of 5 from 10 to 95.

Permutations Of ‘n’ Objects

We have seen what a factorial notation is. Let us say we have ‘n’ objects that we need to arrange in n ways. For example, you have 10 balls and 10 empty spaces to put them into. Something like this. So in how many ways can we do this? Well if you use the fundamental principle of counting, you will say that we can fill the first space in 10 different ways. The second space can be filled in 9 different ways and so on. In other words, the number of permutations we have here will be equal to 10×9×8×7×6×5×4×3×2×1 = 10! We can generalise this result and say that if we have ‘n’ different objects that are to be arranged in n different ways, then we have a total number of n! permutations.

The Case Of Allowing Repetition

What if repetition of the event is allowed? In other words, let us consider this example:  You cast a die twice. How many permutations of the possible outcomes will you get? Let us see for one throw. We can get any of the six outcomes for one throw. In the second throw, we have the same number of outcomes. Therefore from the fundamental principle of counting the number of ways in which we can have the outcomes in two throws of a die = 6×6 = 36.

In general, we say that if we have an event that can happen in ‘n’ ways and is repeated ‘r’ times, then the possible number of outcomes of such an event = nr.

Solved Examples For You

Example 3: How many different car number plates are possible with 3 letters followed by 3 digits?

Answer: The first letters can be chosen from 26 alphabets of the English language and so can the other two. Therefore the arrangements for the first three alphabets or letters are 26×26×26 = (26)3. Similarly, the number of arrangements for the three numbers is = 103. The total number of arrangements or combinations will thus be =  (26)× 103

Example 4: In a city, the bus route numbers consist of a natural number less than 100, followed by one of the letters A, B, C, D, E and F. How many different bus routes are possible?
Answer: The number can be any one of the natural numbers from 1 to 99. There are 99 choices for the number.
The letter can be chosen in 6 ways. ∴ Number of possible bus routes are 99×6 = 594.

Practice Questions

Q 1: In how many ways can 6 people be arranged in a row?

A) 66           B) 72            C) 660              D) 720

Ans: D) 720

Q 2: Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel from A to E?

A) 68             B) 70             C) 72           D) 1024

Ans: C) 72

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2 responses to “Factorial Notation”

  1. Vijayalaxmi says:

    Sir cards probability questions

  2. mk Awas says:

    How to Learn Factorial notation

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