Permutation in Discrete Mathematics- Piyush Bujade

Permutation in Discrete mathematics

1. P e r m u t a t i o n by: Piyush Bujade

2. Definition:- A permutation is a mathematical technique that determines the number of possible arrangements in a finite set when the order of the arrangements matters. Permutation

3. General Formula the total no. of permutations of 'n' distinct objects taken 'r' (r<=n) at a time n! (n-r)! n P(n,r) = P r =

4. A pp l i c a t i on : Emily has 4 chairs and she wants to place 3 dolls on these chairs. In how many possible ways can she do this? Given n = 4 and r = 3 So, there are 24 ways in which she can place 3 dolls on the 4 chairs.

5. Types of Permutation there are two types of Permutation Permutation with Repetition This formula is used to find the statistics of permutation (number of possible ways in which arrangement can be done) while allowing repetition. Permutation without Repetition This formula is used to find the statistics of permutation (number of possible ways in which arrangement can be done) while ensuring that there is no repetition.

6. When repitition is allowed When repitition is not allowed Application of Both Find the number of 3 letter words that can be formed with word "water". n = 5 and r = 3 + + + + 5 4 3 = 60 5 5 5 5 3 = 125

7. Permutation with Non Distinct Objects :- When some objects are alike( Example: word "MISSISSIPPI" where 'I', 'S', 'P' are repeated). If out of 'n' objects in a set 'p' objects are of one kind, 'q' objects are of second kind and 'r' of third then the no. of Arrangements will be- n! p! q! r!

8. The number of permutations of the letters of word ALLAHABAD n=9, A is repeated 4 times and L is 2 times. Rest are of different kind P(n,r) = p! q! n! = 9! 4! 2! = 7560

9. there are two types of Restrictions Permutation with Restrictions (Conditional Permutation) the no. Permutations of 'n' different objects taken 'r' at a time in which 'k' particular objects are Present the no. Permutations of 'n' different objects taken 'r' at a time in which 'k' particular objects do not occur

10. Find the number of ways; the letters of the word ABUJA can be permutated Therefore the number of arrangement is Problem 1 If the two A's must always be apart First Omit out A's, the letter BUJ can be arranged in 3! ways with each of these ways, the first A can be inserted in any one of 4 places * B * U * J * the second A can be inserted in 3 ways, in which it will not come next to the first A but also the two A's cannot be distinguished. Problem 2 If the two A's must always be together Taking the two A's as one object the letter of the word can be arranged in 3! Ways * B * U * J * A's can occupy any of the asterisk positions in 4 ways. Hence the numbers of arrangement is 3! × 4 = 24 ways Or If the 2 A's must always be together, we take them as one then we have (AA) BUJ. The number of permutation of these letters is 4! = 24 ways

11. Thank You!