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July 22, 2016

Permutations & Combinations Problems with Shortcut Methods - Introduction

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Hi friends, today we shall learn the basics of Permutations & Combinations of Aptitude Section. Happy Reading :)   

Factorial Notation
Let n be a positive integer. Then the continued product of first 'n' natural numbers i.e.,
1 x 2 x 3 x 4 x ..... x (n-1) x n is known as 'n' factorial is denoted by n!

Also as per definition 0! = 1

Note : When 'n' is factorial or a negative n! is not defined.

Hence n! = 1 x 2 x 3 x ....... x (n - 2) x (n - 1) x n

Fundamental Principle

Let's consider the following example.
A person wants to travel by a car, bus, train or an aero plane while traveling from Hyderabad to Nagpur. But for the return journey he can travel in any these manners except the aero plane. In how many ways can the man complete his journey ?

Now, the trip from Hyderabad to Nagpur can be done by any one of four means viz. a car, a bus, a train or a plane. For the return journey, only three methods are available. Suppose he goes by car. Then, he can return in three different ways. Hence in this case, the journey can be complete in three ways.

But this is true for every choice of the method to travel from Hyderabad to Nagpur, there are 3 methods available for the return journey.

Hence, corresponding to the 4 ways of gong to Nagpur, he will have 4 x 3 = 12 ways of combining.

We can consider this example as follows :

The man has to perform two actions one after another. The first action ie., going to Nagpur can be performed in 4 different ways and the second action, i.e., returning from Nagpur can be performed in 3 different ways. Hence the total number of ways of performing these two actions together is 4 x 3 = 12.

Multiplication principle 

If one thing can be done in 'm' ways and after it has been done by any one of these 'm' ways another thing can be done in 'n' ways, then the total number of ways of doing the two things is 'm x n'
Permutation 
  1. Each of the arrangement which can be made by taking some or all a number of objects is known as a permutation. 
  2. If there are n different objects and we take r objects out of them at a time, then the number of different arrangements or permutations which can be made are given by 
nPr = n(n - 1)(n - 2) ... (n - r + 1) =n!
(n - r)!

Restricted Permutations

  1. The number of all permutations of n different things taken r at a time, when a particular thing is to be always included in each arrangement is r ( n-1Pr-1 ). 
  2. The number of permutations of n different things taken r at a time, when a particular thing is never taken in each arrangement is n-1Pr.
  3. The number of all permutations of n different things taken all at a time is n! 
Permutations of objects when some of them are identical 
The number of permutations of n objects taken all of them at a time, when P objects are of one kind, q are of a second and r of a third and n! / (p! q! r!) 
Lets have a look at some Examples for better understanding. 

Permutations & Combinations Example Problems with Solutions 
1. Find the number of numbers greater than 3000, that can be formed the digits 1, 2, 3, 4, 5, 6 if repetition is not allowed ?
Solution :
For the number to be greater than 3000, the first digit can assume any value amongst 3, 4, 5, 6;
Provided the number is a 4 digits number.

Hence the first digit can be chose from the unused 5 digits in 5p3 = 60 ways 

=> The answer is 4 x 60 = 240

2. One each of two different types of mangoes and seven different types of apples are to be arranged in a row so that the two mangoes are always together. Find the number of ways this can be done ?
Solution : 
The two mangoes can be arranged among themselves in 2! = 2 ways. Now, these two mangoes are to be together always. So it could be treated as one unit and each apple as one unit each. Hence are 8 units. They can be arranged in 
8! ways = 40320

3. A combination lock using the letters A-Z contains four rings. At the maximum, how many distinct false trails are expected before the lock is opened ?
Solution : 
In this case, each of the rings can attain 26 different positions A to Z. So, the total number of such permutations are 26 x 26 x 26 x 26 = 26^4  = 456976
Of these one is the correct lock combination. 

Hence, the maximum number of distinct false trails that can be made before the lock is opened = 456976 - 1 = 456975

4. A horn can make 9 different sounds. If it is blown 6 times and it should start and end with same sound, in how many ways can it be blown ?
Solution : 
Out of 6 times, the first and last are fixed since they are blown by 9 different sounds with repetition in 9^4 = 6561 ways. 

5. The city of Chennai has seven digit telephone numbers. How many combinations of telephone numbers can exist in which at least one of the digit is repeated ?

Solution : 

Getting at least one of the digits repeated is equal to the difference between the combinations where none repeats and the total possible seven digit telephone numbers. 

Total possible seven digit no's = 90 x 10^6

Number of outcomes in which all digits are different is 

9 x 9 x 8 x 7 x 6 x 5 x 4 = 544320

Number in which at least one of digits is repeated is 

9000000 - 544320 = 8455680

That's all for now friends. In our next post we shall learn about Circular Permutations and the Difference between Permutations and Combinations. Happy Reading :)

shared by Aindree Mukherjee
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