When we look at the schedules of trains, buses and the flights we really wonder how they are scheduled according to the public's convenience. Of course permutation is very much helpful to prepare the schedules on departure and arrival of these. Also when we come across license plates of vehicles which consists of few alphabets and digits. We can easily prepare these codes using permutations.

Apart from this we we can find how many signals can be made with the given flags of different colors, the number of different digit numbers that can be formed from the given digits. In this section let us see about the basic principle called Fundamental Counting principle, Definition of Permutation, its formula, few examples and the practice problems.

According to this principle, "If one operation can be performed in 'm' ways and there are n ways of performing a second operation, then the number of ways of performing the two operations together is m x n ".The above principle can be extended to the case in which the different operation be performed in m, n, p, . . . . . . ways.

In this case the number of ways of performing all the operations one after the other is m x n x p x . . . . . . . . .

Solved Examples

Question 1: How many different numbers of three digits can be formed with the digits 1, 2, 3, 4 and 5 no digit being repeated.
Solution:
 
We have the digits 1, 2, 3, 4 and 5.
Being a three digit number to be formed, it will have places like unit place, ten's place and hundred's place.
Among the given five digits, the unit's place can be filled in 5 ways.

We are left with four digits to fill the ten's place.
Therefore, the ten's place can be filled in 4 ways.

We are left with three digits to fill the hundred's place. Hundred's place can be filled in 3 ways.

Therefore, by the fundamental counting principle, The number of three digit numbers that can be formed is 5 x 4 x 3 = 60.

                   The numbers formed will be like, 123, 124, 125, 132, 134, 135, 142, 143, 145, 152, 153, 154 and so on.
 

Question 2: There are three section in a degree college. Each section has 30 students.
In how many ways three student representatives be selected so that there will be one student from each section.
Solution:
 
There are 30 students in each of three sections.
                                    Let us assume that the three sections are , A, B and C.
                                    1st representative can be selected from Section A in 30 ways.
                                    2nd representative can be selected from Section B in 30 ways.
                                    3rd  representative can be selected from Section C in 30 ways.

Therefore the required number of ways of selecting the representatives = 30 x 30 x 30
                                                                                                                  = 27,000

 

Factorial: The product of first n natural numbers can be denoted in the form n! (or) which is read as " n-Factorial"

For example: 4! = 4. 3. .2. 1 = 24

Example: Find the value of $\frac{6!}{4!}$

Solution:
We have , $\frac{6!}{4!}$ = $\frac{6.5.4.3.2.1}{1.2.3.4}$

= 6.5

= 30

This can also be simplified as $\frac{6!}{4!}$ = $\frac{6.5.4!}{4!}$

= 6.5

$\frac{6!}{4!}$ = 30

Permutations: The different arrangements which can be made out of a given number of things by taking some or all at a time are called permutations.

In a permutation it concerns about the selection as well as arrangement. Ordering is essential in permutations.

The number of permutations of n things taken r at a time will be the same as the number of ways in which r blank places can be filled up with n given things.

When we start filling r different places from n different things, the number of objects will be reduced by 1 as we proceed filling from the first place.

Therefore the first second , third places can be filled in n, (n-1) and (n-2) ways respectively.

Similarly the rth place will be filled in (n - (r-1)) ways = (n - r + 1) ways.

Therefore, by the above Fundamental Principle of Counting, we have
the r places can be filled in n (n-1) (n-2) ...............(n-r+1) ways.
This can be denoted as nPr = n (n-1) (n-2) ...............(n-r+1)

The value of nPr = P (n, r) can be calculated using the formula,
P (n, r) = n . (n-1) . (n-2) ......... (n - r + 1)
In terms of factorial notation, P (n , r) = $\frac{n!}{(n-r)!}$

Case 1: When r = 0, we have nP0 = $\frac{n!}{(n-0)!}$ = $\frac{n!}{n!}$

nP0 = 1

Case 2: When r = 1,we have nP1 = $\frac{n!}{(n-1)!}$ = $\frac{n.\; (n-1)!}{(n-1)!}$

nP1 = n

Case 3: When r = n, we have nPn = $\frac{n!}{(n-n)!}$ = $\frac{n!}{0!}$

nPn = n! [since 0 ! = 1]

0 ! is meaning less, but in order to avoid contradiction in the results, we suppose that 0 ! = 1
The number of permutations of n different things taken r at a time, when each thing may occur any number of times is nr .

Suppose r places are to be filled with n things.
The first place can be filled in n ways.
The second place can also be filled in n ways, since repetition is allowed.
continuing this way, the rth place can also be filled in n ways.
By the Fundamental Counting Principle, the total number of ways to fill in the r places with repetition = n. n. n. . . . . . . r times = nr

Solved Example

Question: In how many ways 6 rings of different type can be had in 4 fingers?
Solution:
 
The first ring can be worn in any of the 4 fingers.
                So, there are 4 ways of wearing it.
                Similarly each one of the other rings may be worn in 4 ways.
                Therefore, Requisite number of ways = 46
                                                                     = 4096

               6 rings of different type can be had in 4096 ways.


 


Solved Examples

Question 1: Evaluate P(6, 4)
Solution:
 
P(6, 4) = 6 (6-1) (6-2) (6-3)
= 6.5.4.3
= 360
 

Question 2: If P(n, 4) = 20 x P(n, 2), find n.
Solution:
 
We have P (n , 4) = 20 P (n , 2)
=> n (n - 1) (n - 2) (n - 3) = 20 x n (n - 1)
=> (n - 2) (n - 3) = 20 [ by dividing both sides by n (n-1) since n $\ne$ 0 and (n-1) $\ne$ 0.
=> n2 - 5n + 6 - 20 = 0
=> n2 - 5n - 14 = 0
=> n2 - 7n + 2n - 14 = 0
=> n (n - 7) + 2 (n - 7) = 0
=> (n - 7) (n + 2) = 0
=> n = -2 or n = 7
Since n cannot be negative, n = 7 which is positive.
 

Question 3: If P(15, r) = 2730, find r.
Solution:
 
We have P(15, r) = 2730
Prime factors of 2730 = 2 x 3 x 5 x 7 x 13 = 13 x 14 x 15 [expressed as product of three consecutive numbers]
Therefore, P(15, r) = 15 x 14 x 13
=> r = 3 [by inspection]
 

Question 4: It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Solution:
 
We are given that there are 5 men and 4 women.
(i. e) there are 9 positions.
The even positions are , 2nd, 4th, 6th and the 8th places
These four places can be occupied by 4 women in P(4, 4) ways = 4 !
= 4 . 3. 2. 1
= 24 ways
The remaining 5 positions can be occupied by 5 men in P(5,5) = 5!
= 5.4.3.2.1
= 120 ways
Therefore, by the Fundamental Counting Principle,
Total number of ways of seating arrangements = 24 x 120
= 2880


 

Question 5: There are 6 English, 5 General Since and 4 Math books. In how many ways can they be arranged on a shelf so as to keep all the books of the same language together?
Solution:
 
Let us make one packet for each of the books on the same language.
Now, 3 packets can be arranged in P (3, 3) = 3 !
= 3 .2 . 1
= 6 ways
6 books on English can be arranged in 6 ! = 1 . 2 . 3 . 4 . 5 . 6
= 720 ways
5 books on General Science can be arranged in 5 ! = 1 . 2 . 3 . 4 . 5 = 120 ways
4 math books can be arranged in 4! ways = 1 . 2 . 3 . 4 = 24 ways
therefore, the Required number of ways = (6 x 720 x 120 x 24) ways
 


Practice Problems

Question 1: sdfsdf
Question 2: Prove that P(n, n) = 2 P (n, n-2).
Question 3: If 9P5 + 5 . 9P4 = 10 Pr , find r.
Question 4: Seven athletes are participating in a race. In how many ways can the first three prizes be won.
Question 5: How many numbers lying between 100 and 1000 can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed.
Question 6: A license plate consists of two distinct English alphabet followed by two distinct numbers from 1 to 9. How many such plates can be made?