# Permutations, number of arrangements without elements repetition

### Permutation:

An arrangement that can be formed by taking some or all of a finite set of things (or objects) is called a permutation.
Order of the things is very important in case of permutation.
A permutation is said to be a Linear Permutation, if the objects are arranged in a line. A linear permutation is simply called as a permutation.
A permutation is said to be a Circular Permutation, if the objects are arranged in the form of a circle.
The number of (linear) permutations that can be formed by taking r things at a time from a set of n distinct things is denoted nPr or P(n;r).

nPr=n(n-1)(n-2)(n-3)×…×(n-r+1) ### Permutations when all the objects are distinct

Theorem 1: The number of permutations of n different objects taken r at a time, where 0<r≤n and the objects do not repeat is n(n-1)(n-2)…(n-r+1), which is denoted by nPr.
Proof: There will be as many permutations as there are ways of filling in r vacant places

[_][_][_]…[_]
–>r vacant places<–

Then n objects. The first place can be filled in n ways; following which, the second place can be filled in (n-1) ways, following which the third place can be filled in (n-2) ways,…, the rth place can be filled in (n-(r-1)) ways. Therefore, the number of ways of filling in r vacant places in succession is n(n-1)(n-2)…(n-r+1).
This expression for nPr is cumbersome and we need a notation which will help to reduce the size of this expression. The symbol r! (read as factorial n or n factorial) comes to our rescue. In the following text we will learn what actually n! means.
7.3.2 Factorial notation The notation n! represents the product of first n natural numbers, i.e., the product 1×2×3×…×(n-1)×n is denoted as n!. We read this
symbol as ‘n factorial’. Thus, 1×2×3×…×(n-1)×n=n!
1=1!
1×2=2!
1×2×3=3!
1×2×3×4=4! and so on.
We define 0!=1
We can write 5!=5∙4!=5∙4∙3!=5∙4∙3∙2!=5∙4∙3∙2∙1!
Clearly, for a natural number n
n!=n(n-1)!
=n(n-1)(n-2)! [provided (n≥2)]
=n(n-1)(n-2)(n-3)! [provided (n≥3)]
and so on.

Example of theorem 1: Evaluate (i). 5! (ii). 7! (iii). 7!-5!
solution: (i). 5!=1×2×3×4×5=120
(ii). 7!=1×2×3×4×5×6×7=5040
and (iii). 7!-5!=5040-120=4920.

Theorem 2: The number of permutations of n different objects taken r at a time, where repetition is allowed, is n^r.
Proof is very similar to that of Theorem 1 and is left for the reader to arrive at.
Here, we are solving some of the problems of the pervious Section using the formula for nPr to illustrate its usefulness.
In Example 1, the required number of words =4P4=4!=24. Here repetition is not allowed. If repetition is allowed, the required number of words would be 4^4=256.
The number of 3-letter words which can be formed by the letters of the word NUMBER =6P3=6!/3!=4×5×6=120. Here, in this case also, the repetition is not allowed. If the repetition is allowed, the required number of words would be 6^3=216.
The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly =12P2=12!/10!=11×12=132.

Example of theorem 2: How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
solution: Here order matters for example 1234 and 1324 are two different numbers. Therefore, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
Therefore, the required 4 digit numbers ### Simple questions and solutions about pemutation

Q1. How many 3-digit numbers can be formed by using the digits 2 to 9 if no digits is repeated?
solution:
3-digit numbers have to be formed using the digits 1 to 9.
Here, the order of the digits matters.
Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.
Therefore, required number of 3-digit numbers equals Q2. How many 3-digit even numbers can be formed from the given six digits 1, 2, 3, 4, 6 and 7, without repeating the digits?
solution:
3-digit even numbers are to be formed using the given six digits, 1, 2, 3, 4, 6, and 7, without repeating the digits.
Then, units can be filled in 3 ways by any of the digits, 2, 4 or 6.
Since the digits cannot be repeated in the 3-digit numbers and units place is already seated
with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5 digits.
Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits is the permutation of 5 different digits taken 2 at a time.
Number of ways of filling hundreds and tens place equals Thus, by multiplication principle, the required number of 3-digit numbers is 3×20=60.

Q3. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, if no digit is repeated. How many of these will be even?
solution: Q4. How many ways can be made for choosing a chairman and a vice chairman from a committee of 8 persons?
solution:
From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such as way that one person cannot hold more than one position.
Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8
different objects taken 2 at a time.
Thus, required number of ways equals Q5. How many words, with or without meaning, can be formed using all the letters of the word =EQUATION, using each letter exactly once?
solution:
There are 8 different letters in the word EQUATION.
Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is the number of permutations of 8 different objects taken 8 at a time, which is 9P8=8!.
Thus, required number of words that can be formed =8!=40320.

Q6. In how many ways can a three-letter word be made from the letters c; d; e; f without repeating any letters?
solution:
4P3=4!/(4-3)!=4!/1!=24 ways, a 3-letter word can be made from c; d; e; f with no repetition.

Q7. How many signals can be given by 5 flags of different colours, using 3 at a time?.
solution:
Given that n=5,r=3
Required numbers of signals =5P3 Q8. How many signals can be given by 6 flags of different colours when any number of flags can
be used at a time?
solution: Number of signals using 1 flags =6P1=6
Number of signals using 2 flags =6P2=30
Number of signals using 3 flags =6P3=120
Number of signals using 4 flags =6P4=360
Number of signals using 5 flags =6P5=720
Number of signalg using 6 flags =6P6=720
Required number of signals

=6+30+120+360+720+720=1956
Q9. How many 3-digit numbers can be formed by using each one of the digits 2,3,5,7,9 only once?
solution:
Given that n=5,r=3
Required numbers =5P3 Q10. Find the numbers greater than 23000 that can be formed from the digits 1,2,3,5,6 without repeating any digit.
solution: Numbers greater than 23000 are of the form
Numbers with 23 on the extreme left 23000 =3P3=6
Numbers with 25 on the extreme left 25000 =3P3=6
Numbers with 26 on the extreme left 26000 =3P3=6
Numbers with 3 on the extreme left 30000 =4P4=24
Numbers with 5 on the extreme left 50000 =4P4=24
Numbers with 6 on the extreme left 60000 =4P4=24
Hence total numbers are: 6+6+6+24+24+24=90

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