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Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams PDF Download

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:

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

= 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:

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

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 .

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

 

Ques 7. How many even numbers of three digits can be formed with the digits 0,1, 2, 3, 4, 5 and 6 ?

Sol :

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

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:

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

1. Factorial Notation:

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n – 1)(n – 2) … 3.2.1.

Examples:

  1. We define 0! = 1.
  2. 4! = (4 x 3 x 2 x 1) = 24.
  3. 5! = (5 x 4 x 3 x 2 x 1) = 120.

 2. Permutations:

The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Examples:

  1. All permutations (or arrangements) made with the letters abc by taking two at a time are (abbaaccabccb).
  2. All permutations made with the letters abc taking all at a time are:
    ( abcacbbacbcacabcba)

3. Number of Permutations:

Number of all permutations of n things, taken r at a time, is given by:

nPrn(n – 1)(n – 2) … (n – r + 1) = Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

Examples:

  1. 6P2 = (6 x 5) = 30.
  2. 7P3 = (7 x 6 x 5) = 210.
  3. Cor. number of all permutations of n things, taken all at a time = n!.

4. An Important Result:

If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind;p3 are alike of third kind and so on and pr are alike of rth kind, 
such that (p1p2 + … pr) = n.

Then, number of permutations of these n objects is = Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

5. Combinations:

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Examples:

  1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.Note: AB and BA represent the same selection.
  2. All the combinations formed by abc taking abbcca.
  3. The only combination that can be formed of three letters abc taken all at a time is abc.
  4. Various groups of 2 out of four persons A, B, C, D are:

    AB, AC, AD, BC, BD, CD.

  5. Note that ab ba are two different permutations but they represent the same combination.

6. Number of Combinations:

The number of all combinations of n things, taken r at a time is:

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

Note:

  1. nCn = 1 and nC0 = 1.
  2. nCrnC(n – r)

Examples:

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams

The document Permutation and Combination - Introduction and Examples (Part - 2) | Quantitative Aptitude for Competitive Examinations - Banking Exams is a part of the Banking Exams Course Quantitative Aptitude for Competitive Examinations.
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FAQs on Permutation and Combination - Introduction and Examples (Part - 2) - Quantitative Aptitude for Competitive Examinations - Banking Exams

1. What is the difference between permutation and combination?
Ans. Permutation is an arrangement of objects in a specific order, while combination is a selection of objects without considering the order. In permutation, the order matters, but in combination, it doesn't.
2. How do you calculate the number of permutations of a set?
Ans. The number of permutations of a set can be calculated using the formula nPr = n! / (n-r)!, where n is the total number of objects and r is the number of objects taken at a time.
3. Can you explain the concept of factorial in permutation and combination?
Ans. Factorial is a mathematical function denoted by the symbol "!". In permutation and combination, factorial is used to calculate the number of ways objects can be arranged or selected. It represents the product of all positive integers less than or equal to a given number.
4. What is the difference between a permutation and a combination in real-life examples?
Ans. In real-life examples, a permutation represents the different ways in which a group of people can be arranged in a line or seated at a table, where the order matters. On the other hand, a combination represents the different groups that can be formed by selecting a certain number of people from a larger group, where the order doesn't matter.
5. How can permutation and combination be applied in probability problems?
Ans. Permutation and combination are used in probability problems to calculate the number of favorable outcomes and the total number of possible outcomes. By dividing the number of favorable outcomes by the total number of possible outcomes, we can determine the probability of an event occurring.
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