Permutation and Combination Tricks with Questions - PDF

Permutation and Combination
Today, I am going to share techniques to solve  permutation and combination questions. This chapter talk about selection and arrangement of things which could be any numbers, persons,letters,alphabets,colors etc.The basic difference between permutation and combination is of order.



Permutation

Permutation is basically called as a arrangement where order does matters.Here we need to arrange the digits , numbers , alphabets, colors and letters taking some or all at a time.It is represented as nPr .
Permutation

Permutation
Note :
1. nPr = n! / (n-r)!
2. If from the total set of n numbers p is of one kind and q ,r are others respectively then nPr = n! / p! × q! × r!.
3. nPn = n!

Combination

Combination is basically called as a selection where order does not matters.Here we need to arrange the digits , numbers , alphabets, colors and letters taking some or all at a time. It is represented as nCr.

combinations

combinations
Note :
  1.  nCr = n!/ r! × (n-r)!  
  2.  nC0 = 1
  3.  nCn = 1
  4.  nCr =  nCn - r 
  5.  nCa =  nCb => a = b => a+b = n
  6. nC0nC1nC2nC3+ ...............+ nC= 2n

Permutation vs Combination

In both the things main difference is of order .In permutation order matters while in combination it does not.
Basic Difference :
  1. order
  2. arrange or choose 
  3. number of permutation > number of combination
Permutation vs Combination

Permutation vs Combination


Permutation vs Combination
scene
In Arrangements we have,
Total no. of arrangements = total no. of groups or selection × r!
where r is the no. of objects in each group or selection. So nPr = nCr  × r!

Questions :

1. How many triangles can be formed with four points (A,B,C & D) in a plane ? It is given that no three points are col-linear(not comes in straight line).From the three points A,B and C have only one triangle with these points.
Sol:
Here in this question  , Order of digit does not matter so it is a combination.
 nCr = n!/ r! × (n-r)!  
 4C= 4!/3! × ( 4- 3)! 
 4C= 4!/3! × 1! 
 4C= 4!/3! × 1!
 4C= 4
Or
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2. How many number plates of 3 digit can be formed with four digits 1,2,3 and 4 ?
Sol:
Here, the order of arrangement of digits does matter.
nPr = n! / (n-r)!
nPr = 4! / (4-3)!
4P3 = 4! / 1!
 4P3 = 4!
 4P3 = 24

Factorial Notation

To solve problem like this you must have the knowledge of factorial.Factorial is represented as like " ! ".The Factorial notation is :
questions

Counting Rules

Multiplication

Suppose one starts his journey from place X and has to reach place Z via a different place Y. For Y, there are three means of transport - bus, train and aeroplane - from X. From Y, the aeroplane service is not available for Z. Only either by a bus or by a train can one reach Z from Y. Also, there is no direct bus or train service fro Z from X. We want to know the maximum possible no. of ways by which one can reach Z from X.
Sol:
questions
= 3  × 2
= 6
If a work A can be done in m ways and another work B can be done in n ways and C is the final work which is done only when both A and B are done, then the no. of ways of doing the final work ?
Sol :
C = m  × n
C = 3 × 2 = 6

Addition rule

Suppose there are 42 men and 16 women in a party. Each man shakes his hand only with all the men and each woman shakes her hand only with all the women. We have to find the maximum no. of handsakes that have taken place at the party.
From each group of two persons we have one handshake.
Case 1 : Total no. of handshakes among the group of 42 men
42C= 42!/2! (42-2)! = 21 × 41 = 861
Case 2 : Total no. of  handshakes among the group of 16 women
16C= 16!/2! (16-2)! = 8 × 15 = 120
so maximum no. of handshakes = 861 + 120 = 981.

Problems and Solutions

Ques 1. How many numbers of five digits can be formed with the digits 1,3,5 7 and 9 no digit being repeated ?Sol :
the no. of digits = 5
Required no. = 5P5 = 5! = 120

Ques 2. How many three-digit numbers can be formed by using the digits in 735621, if repetition is not allowed ?
Sol:
nPr = n! / (n-r)!
6P=  6! / (6-3)!
6P= 6!/3!
6P= 120

Ques 3. Find the number of different words that can be formed from the word 'SUCCESS'.
Sol : No. of Permutation = n! / p! × q!, where p = of one type , q = ( of another type ).
No. of Permutation = 7!/ 3! × 2!
No. of Permutation = 420

Ques 4. How many different 5 - digit numbers can be formed by using the digits of the number 713628459 ?
Sol :
nPr = n! / (n-r)!
9P5 = 9! / (9-5)!
9P5 = 9! / 4!
9P5 = 15,120 

Ques 5. How many numbers of five digits can be formed with the digits 0,2,4,6 and 8 ?
Sol:
questions

Ques 6. How many numbers of five digits can be formed with the digits 0,1, 2, 3, 4, 6 and 8 ?
Sol :
Here nothing has been said about the repetition of digits. So , it is understood that repetition of digits is not allowed .
questions
Ques 7. How many even numbers of three digits can be formed with the digits 0,1, 2, 3, 4, 5 and 6 ?
Sol :
questions
Total of such numbers = 5 × 5 v 3 = 75
req no. = 30+75 = 105

Ques 8. A round table conference is to be held between delegates of 15 companies. In how many ways can they be seated if delegates from two MNCs may wish to sit together ?
Sol :
Since delegates from two multinational companies will sit together, so considering these two delegates as one unit, there will be 13 + 1 = 14 delegates who can be arranged in a circular table in 14! ways.
The two delegates from the MNCs can be arranged among themselves in 2! ways.
Using the product rule, the required no. of ways = 14!×2!

Ques 9. A person has 12 friends out of which 7 are relatives. In how many ways can he invite 6 friends such that at least 4 of them are relatives ?
Sol:
questions

Question 10. How many numbers of five digits can be formed with the digits 0, 1, 2, 4, 6 and 8 ?

Sol:
Required no. of numbers = 5 × 5P4= 5 × 5! = 5×120 = 600

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