Permutations And Combinations

MADE BY :- ANUBHAV KUMAR

CLASS :- 11TH ‘A’

Introduction

Suppose you have a suitcase with a number lock. The number lock has 4 wheels each labelled with10 digits from 0 to 9. The lock can be opened if 4 specific digits are arranged in a particularsequence with no repetition. Some how, you have forgotten this specific sequence of digits. Youremember only the first digit which is 7. In order to open the lock, how many sequences of 3-digitsyou may have to check with? To answer this question, you may, immediately, start listing all possiblearrangements of 9 remaining digits taken 3 at a time. But, this method will be tedious, because thenumber of possible sequences may be large. Here, in this Chapter, we shall learn some basiccounting techniques which will enable us to answer this question without actually listing 3-digitarrangements. In fact, these techniques will be useful in determining the number of different ways ofarranging and selecting objects without actually listing them. As a first step, we shall examine aprinciple which is most fundamental to the learning of these techniques.

Jacob Bernoulli

Jacob Bernoulli (also known as James or Jacques; 6 January 1655 [O.S. 27 December 1654] – 16 August 1705) was one of the many prominent mathematicians in the Bernoulli family. He was an early proponent of Leibnizian calculus and had sided with Leibniz during the Leibniz–Newton calculus controversy. He is known for his numerous contributions to calculus, and along with his brother Johann, was one of the founders of the calculus of variations. However, his most important contribution was in the field of probability, where he derived the first version of the law of large numbers in his work Ars Conjectandi.

OVERVIEW

The study of permutations and combinations is concerned with determiningthe number of different ways of arranging and selecting objects out of agiven number of objects without actually listing them. There are some basiccounting techniques which will be useful in determining the number ofdifferent ways of arranging or selecting objects. The two basic countingprinciple are given below:

FUNDAMENTAL PRINCIPLE OF COUNTING

MULTIPLICATION PRINCIPLE

Suppose an event E can occur in m different waysand associated with each way of occurring of E,another event F can occur in n different ways, thenthe total number of occurrence of the two events inthe given order is m × n.

ADDITION PRINCIPLE

If an event E can occur in m ways and anotherevent F can occur in n ways, and suppose thatboth can not occur together, then E or F canoccur in m + n ways.

Permutations

A permutation is an arrangement of objects in a definite order.

Permutation of n different objects

The number of permutations of n objects taken all at a time, denoted by the symbol nPn , is given by

nPn = n …………. (1)

where n = n(n – 1) (n – 2) ... 3.2.1, read as factorial n, or n factorial. The number of permutations of n objects taken r at a time, where 0 < r ≤ n, denoted by nPr , is given by

nPr = n / n-r

We assume that 0 = 1

When repetition of objects is allowed

The number of permutations of n thingstaken all at a time, when repetion of objectsis allowed is nn. The number of permutationsof n objects, taken r at a time, whenrepetition of objects is allowed, is nr. 7.1.6

Permutations when the objects are not distinct

The number of permutations of n objects ofwhich p1 are of one kind, p2 are of second kind,..., pk are of kth kind and the rest if any, are ofdifferent kinds is

n!/p1! p2! ……pk!

Combinations

On many occasions we are not interested in arranging butonly in selecting r objects from given n objects. A combinationis a selection of some or all of a number of different objectswhere the order of selection is immaterial. The number ofselections of r objects from the given n objects is denoted bynCr , and is given by

nCr = n!/ r! (n-r)!

Remarks

1. Use permutations if a problem calls for the number ofarrangements of objects and different orders are to becounted.

2. Use combinations if a problem calls for the number of waysof selecting objects and the order of selection is not to becounted.

Some important results

Let n and r be positive integers such that r ≤ n. Then

(i) nCr = nCn – r

(ii) nCr + nCr – 1 = n + 1Cr

(iii) n n – 1Cr – 1 = (n – r + 1) nCr –1

Example 1

In a class, there are 27 boys and 14 girls. The teacher wants to select 1 boy and 1 girl torepresent the class for a function. In how many ways can the teacher make this selection?

Solution

Here the teacher is to perform two operations:

(i) Selecting a boy from among the 27 boys and

(ii) Selecting a girl from among 14 girls.

The first of these can be done in 27 ways and second can be performed in 14 ways. Bythe fundamental principle of counting, the required number of ways is 27 × 14 = 378.

Example 2

(i) How many numbers are there between 99 and 1000 having 7 in the units place?

(ii) How many numbers are there between 99 and 1000 having at least one of their digits 7?

Solution

(i) First note that all these numbers have three digits. 7 is in the unit’s place. The middle digit can beany one of the 10 digits from 0 to 9. The digit in hundred’s place can be any one of the 9 digits from1 to 9. Therefore, by the fundamental principle of counting, there are 10 × 9 = 90 numbers between99 and 1000 having 7 in the unit’s place.

(ii) Total number of 3 digit numbers having at least one of their digits as 7 = (Total numbers of threedigit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all). = (9 × 10 × 10)– (8 × 9 × 9) = 900 – 648 = 252.

Example 11

A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Mathematics Part II, unless Mathematics Part I is also borrowed. In how many ways can he choose the three books to be borrowed?

Solution

Let us make the following cases:

Case (i) Boy borrows Mathematics Part II, then he borrows Mathematics Part I also. So the number of possible choices is 6C1 = 6. Case

(ii) Boy does not borrow Mathematics Part II, then the number of possible choices is 7C3 = 35. Hence, the total number of possible choices is 35 + 6 = 41.