The document Permutation and Combination - Introduction and Examples, Quantitative Aptitude (Part - 2) Quant Notes | EduRev is a part of the Quant Course Quantitative Aptitude for Competitive Examinations.

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**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:**

= 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

From each group of two persons we have one handshake.

**Case 1 : **Total no. of handshakes among the group of 42 men

^{42}C_{2 }= 42!/2! (42-2)! = 21 × 41 = 861

**Case 2 : **Total no. of handshakes among the group of 16 women

^{16}C_{2 }= 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. = ^{5}P_{5} = 5! = 120

**Ques 2. How many three-digit numbers can be formed by using the digits in 735621, if repetition is not allowed ?**

**Sol:**

^{n}P_{r} = n! / (n-r)!

^{6}P_{3 }= 6! / (6-3)!

^{6}P_{3 }= 6!/3!

^{6}P_{3 }= 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 :**

^{n}P_{r} = n! / (n-r)!

^{9}P_{5} = 9! / (9-5)!

^{9}P_{5} = 9! / 4!

^{9}P_{5} = 15,120

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

**Sol:**

**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 .

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

**Sol :**

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:**

**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:*

- We define
*0! = 1*. - 4! = (4 x 3 x 2 x 1) = 24.
- 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:*

- All permutations (or arrangements) made with the letters
*a*,*b*,*c*by taking two at a time are ().*ab*,*ba*,*ac*,*ca*,*bc*,*cb* - All permutations made with the letters
*a*,*b*,*c*taking all at a time are:

(*abc*,*acb*,*bac*,*bca*,*cab*,*cba*

**3. Number of Permutations:**

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

^{n}P_{r} = *n*(*n* – 1)(*n* – 2) … (*n* – *r* + 1) =

*Examples:*

^{6}P_{2}= (6 x 5) = 30.^{7}P_{3}= (7 x 6 x 5) = 210.*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 *p*_{1} are alike of one kind; *p*_{2} are alike of another kind;*p*_{3} are alike of third kind and so on and *p*_{r} are alike of *r*^{th} kind,

such that (*p*_{1} + *p*_{2} + … *p*_{r}) = *n*.

Then, number of permutations of these *n* objects is =

**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:*

- 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.
- All the combinations formed by
*a*,*b*,*c*taking.*ab*,*bc*,*ca* - The only combination that can be formed of three letters
*a*,*b*,*c*taken all at a time is.*abc* - Various groups of 2 out of four persons A, B, C, D are:
*AB, AC, AD, BC, BD, CD*. - 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:

*Note:*

^{n}C_{n}= 1 and^{n}C_{0}= 1.^{n}C_{r}=^{n}C_{(n – r)}

*Examples:*

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