**MATHS:: Lecture 16 :: PERMUTATION AND COMBINATION**

** PERMUTATION AND COMBINATION**

**Fundamental Counting Principle**

If a first job can be done in ** m** ways and a second job can be done in

**ways then the total number of ways in which both the jobs can be done in succession is**

*n*

*m x n.*For example, consider 3 cities Coimbatore, Chennai and Hyderabad. Assume that there are 3 routes (by road) from Coimbatore to Chennai and 4 routes from Chennai to Hyderabad. Then the total number of routes from Coimbatore to Hyderabad via Chennai is 3 x 4 =12. This can be explained as follows.

For every route from Coimbatore to Chennai there are 4 routes from Chennai to Hyderabad. Since there are 3 road routes from Coimbatore to Chennai, the total number of routes

**is 3 x 4 =12.**

**The above principle can be extended as follows**

**.**If there are

*n*jobs and if there are

*mi*ways in which the ith job can be done, then the total number of ways in which all the

*n*jobs can be done in succession ( 1st job, 2nd job, 3rd job…

*n*th job) is given by

*m1 x m2 x m3 …x mn .*

**Permutation**

**Permutation means**

*arrangement*of things. The word

*arrangement*is used, if the order of things is considered. Let us assume that there are 3 plants P1, P2, P3. These 3 plants can be planted in the following 6 ways namely

P1 |
P2 |
P3 |

P1 |
P3 |
P2 |

P2 |
P1 |
P3 |

P2 |
P3 |
P1 |

P3 |
P1 |
P2 |

P3 |
P2 |
P1 |

Each arrangement is called a permutation. Thus there are 6 arrangements (permutations) of 3 plants taking all the 3 plants at a time. This we write as 3P3. Therefore 3P3 = 6. Suppose out of the 3 objects we choose only 2 objects and arrange them. How many arrangements are possible? For this consider 2 boxes as shown in figure.

I Box |
II Box |

Permutation

Since we want to arrange only two objects and we have totally 3 objects, the first box can be filled by any one of the 3 objects, (i.e.) the first box can be filled in 3 ways. After filling the first box we are left with only 2 objects and the second box can be filled by any one of these two objects. Therefore from Fundamental Counting Principle the total number of ways in which both the boxes can be filled is 3 x 2 =6. This we write as 3*P*2 = 6.

In general the number of permutations of *n* objects taking *r* objects at a time is denoted by *nPr*. Its value is given by

Note: 1

- nPn = n ! (b ) nP1= n. (c) nP0= 1.

**Examples:**

1. Evaluate *8P3 *

**Solution:**

2. Evaluate 11P2

**Solution:**

3. There are 6 varieties on brinjal, in how many ways these can be arranged in 6 plots which are in a line?

**Solution**

Six varieties of brinjal can be arranged in 6 plots in *6P6* ways.

*6*P*6* = = 6! [0! = 1]

= 6 x 5 x 4 x 3 x 2x1 = 720.

Therefore 6 varieties of brinjal can be arranged in 720 ways.

4. There are 5 varieties of roses and 2 varieties of jasmine to be arranged in a row, for a photograph. In how many ways can they be arranged, if

**(i) all varieties of jasmine together **

** (ii)All varieties of jasmine are not together.**

**Solution**

i) Since the 2 varieties of jasmine are inseparable, consider them as one single unit. This together with 5 varieties of roses make 6 units which can be arranged themselves in 6! ways.

In every one of these permutations, 2 varieties of jasmine can be rearranged among themselves in 2! ways.

Hence the total number of arrangements required

= 6! x 2! = 720 x 2 = 1440.

ii)The number of arrangements of all 7 varieties without any restrictions =7! = 5040

Number of arrangements in which all varieties of jasmine are together = 1440.

Therefore number of arrangements required = 5040 -1440 = 3600.

**Combinatination**

Combination means *selection* of things. The word *selection* is used, when the order of thing is immaterial. Let us consider 3 plant varieties V1, V2 & V3. In how many ways 2 varieties can be selected? The possible selections are

1) |
V1 |
& |
V2 |

2) |
V2 |
& |
V3 |

3) |
V1 |
& |
V3 |

Each such selection is known as a combination. There are 3 selections possible from a total of 3 objects taking 2 objects at a time and we write 3C2= 3.

In general the number of selections (Combinations) from a total of *n* objects taking *r* objects at a time is denoted by *n Cr.*

Combination

Probability using Combinations

**Relation between nPr and nCr**

We know that

*n Pr = nCr*x

*r!*

(or) ------------(1)

But we know ---------(2)

Sub (2) in (1) we get

**Another formula for**

*nCr*We know that

*nPr = n. (n-1). (n-2)…(n-r+1)*

\

**Example **

1. Find the value of 10*C*3*.*

**Solution:**

**Note -1**

*nC0 = 1**nC1 = n**nCn = 1*

* d) nCr = nCn-r*

**Examples**

1**. Find the value of 20 C18**

**Solution**

We have 20

*C*18

*=*20

*C*20-18

*=*20C2 = =190

**2.**

**How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?**

**Solution**

Any one prize can be given to any one of the 3 boys and hence there are 3 ways of distributing each prize.

Hence, the 4 prizes can be distributed in 34= 81 ways.

**3. A team of 8 students goes on an excursion, in two cars, of which one can accommodate 5 and the other only 4. In how many ways can they travel?**

**Solution**

There are 8 students and the maximum number of students can accommodate in two cars together is 9.

We may divide the 8 students as follows

**Case I: 5 students in the first car and 3 in the second**

Case II: 4 students in the first car and 4 in the second

In Case I:8 students are divided into groups of 5 and 3 in 8C3 ways.

Case II: 4 students in the first car and 4 in the second

In Case I:

**Similarly, in Case II:**8 students are divided into two groups of 4 and 4 in 8C4 ways.

Therefore, the total number of ways in which 8 students can travel is 8C3 + 8C4 = 56 + 70 = 126.

**4.**

**How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?**

**Solution**

4 consonants out of 12 can be selected in 12C4 ways.

3 vowels can be selected in 4C3 ways.

Therefore, total number of groups each containing 4 consonants and 3 vowels

= 12C4 * 4C3

Each group contains 7 letters, which can be arranging in 7! ways.

Therefore required number of words = 124 * 4C3 * 7!

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