Permutation and Combination Part- 1 by K. Bhanumoorthy

 

 Permutation and combination.

 

Mathematics is not about numbers, equations, computations, or algorithms: it is about understanding.” – William Paul Thurston

 

Let me discuss the very meaning of the word, Permutation. The word is derived from the Latin word ‘ permutare ’. With reference to the Dictionary ‘permutate’ means alter the sequence of arrangement’. Permutation is an ordered arrangement or grouping of a set of numbers /items/ things. Here the order by which arrangement done is important.

To recall the term factorial and the notation as we are using this notation. The continued product of the first n natural numbers is called n factorial and denoted by n! This factorial n is defined with respect to the whole numbers only, not defined for fractions and negative integers. This is to be noted.

Eg : 5 ! = 1x2x3x4x5 = 120. This 5! can also be written as 5 (4!); 5.4.(3!), so on.

 

 There are two principles for counting.

i)                   Principle of addition: If two events are independent, performed one in ‘m ‘ways and the another in ‘n ‘ways, then either of two is carried out in m + n ways.

ii)                 Principle of multiplication: If one of the events is done in ‘m’ ways, after completion, second event is carried out in ‘n ‘ways then two events can be completed in m x n = ways.

Notation: P (n, r) means taking ‘r; things at a time from “n’ things.

The formula given is: P (n, r) = n! / (n-r)!  P (n, n) = n!  To note 0! = 1

If there are certain things are repeated, there is a slight change in the

formula. Eg : If ‘m ‘ things of one kind and ‘n’ things of another kind then

P(n,r) = n! / ( m!.n!)

 

 

 

Application of permutation and combination in daily Life.

1.In a queue, there is 1^st,2^nd,3^rd,4^th, so on. We call them they are in ordinal numbers. They can be arranged in different forms/ways.

2, Number lock, permutating numbers. It depends on one’s choice.

3. In license plate two wheelers or 3wheelers or 4 wheelers we see the arrangement of digits.

In our country India, for a car, the first two letters denote the state code, then with the concerned RTO number where the vehicle is registered. Further prefixing 4 digits, there are any one or two English alphabet from A-to-Z repetition is allowed.

Some people may like to have their preference numbers such as sum has to be 10/ all are odd/all to be even/, to be divisible by 3/ all primes. One can think of in how many ways a number can have 4 digits from 0 to 9 with repetition, the serial numbers from 0001 to 9999. In this case also repetition shall be there. This kind of numbering and registration will differ from country to country.

4.Selecting nominees for students forum, selecting a cricket team, football team etc.

5.Communication networks, cryptography (to say routing different permutations.)

There are questions only on permutation, some on  combination, some combining both.

  

Let us discuss some problems: (Permutation)

1.     Find the number of arrangements of the letters of the word ‘BANANA’ in which two ‘N’ do not appear adjacently.

 Ans: There are 6 letters in 3Asand 2 Ns and 1 B. This can be arranged in 6! / 3! 2!       After simplification we get 60.

The number of arrangements when 2 Ns together is 5! / 3! = 20.
hence the answer is 60 – 20 = 40 where 2 Ns do not appear together.

2.     A five-digit number divisible by 3 is to be formed using the digits 0,1,2,3,4,5 without repetition. Find the total number of ways, this can be done.

 Ans: It is to be a 5-digit number; hence 1^st place cannot be 0 and since divisible by 3, the sum of digits has to be divisible by 3.

If we take 1,2,3,4,5, can be arranged in 5! Ways ie 120 ways.  ---(i), if we take 0,1,2,4,5 the number of 5 digits shall be 4x4x3x2 = 96 ways (5! – 4!)- ----(ii)

Hence the answer is 120 + 96 = 216.

3.Find the number of ways in which we can get a score of 11 by throwing three dice.

Ans: The possible groups for total of 11 are (6,1,4), (6,2,3), (5,2,4), (5,5.1),5,3,3), (4,4,3).  Total 6 groups in which first 3 groups have different numbers and in second lot out of 6, in 3 groups, certain digits are repeated. applying principle of permutation 

We get 3! + 3! + 3! + 3! / 2! + 3! /2! +3! /2! = 6+6+6+3+3+3 = 27.

The answer is 27 ways we can get total of 11.

4.How many three digit odd numbers can be formed using the digits 1,2,3,4,5,6. Repetition is not allowed.

Ans : Since to form odd numbers unit’s place has to be 1,3 or 5.this can be filled in 3 ways. Now ten’s place can be filled by any one of remaining 5 digits, now hundred’s place. This can be filled with any one of 4 digits, be done 4 ways. Hence the answer is 3 x5 x 4 = 60 ways.

5.In how many ways can 6 persons be seated in a row?

Ans: There are 6 persons, they can be arranged in 6 different ways, hence the answer is   6! i.e. 720 ways.

6. How many 9-digit numbers of different digits can be formed?

Ans: Since 9-digit number, first place cannot be 0(zero), remaining 9 digits can be arranged in 9! ways.

Hence the answer is 9. (9!  ways.

 

7. There are 10 lamps in a hall, each one of them can be switched on  

    independently. Find the number of ways, the hall can be illuminated?

  Ans: Let us remember every switch has two options either on or off.

   Hence the answer is 2^10 – 1.

8.Find the number of 5 -letter words that can be formed from the 26 English   alphabet such that the first and the last letters are distinct vowels and remaining are distinct consonants.

Ans: There are 5 vowels ,2 vowels from 5 vowels in 2 extreme positions can be arranged in P (5,2) ways i.e.  60 ways. The remaining 3 places from 21 consonants be arranged in, P (21,3) ways.

Hence the answer is: P (5,2) x P (21,3) ways.

9.If P (56, r+6): P (54, r+3) = 30800: 1, find r?  Nr. 56! / (50-r)!).; Dr. 54! / (51-r)!

  Ans: { 56! / (50-r)!)} / { (54! / (51 – r)! } = 30800

  After cancelling, simplifying, we get, 56.55.(51-r) = 11x 14x 10 x 20,

  51 -r = 10, so r = 41 

10.In a college of 300 students, every student reads 5 newspapers and every 

       newspaper is read by 60 students. Find the number of newspapers.

Ans : Let the number of newspapers be n .  60n = 300x 5, hence n = 25.

          Number of newspapers 25.

 

                                                                                ------- be continued Combination

The author is a retired Principal of Kendriya Vidyalaya.