1. Permutations With Repetitions By Mr Porter 9 Maths 8 Physics7 Biology8 Physics 7 Biology DEFIONTNII CLGEBRAAI

2. Arrangement with Repetition. The number of arrangement of n-elements in a row, when there are ‘p’ alike of one element, ‘q’ alike of another is: The number of permutations of n-elements in a row, chosen r times, with replacement, is given by

3. Example 1 How many different car number-plates can be made if each number-plate contains 4 upper case letters from the alphabet (26) followed by 2 digits from 0 to 9? letters digits In this case, letters and digits can be repeated: AAAA 00ABBA 99XXPP 22….. Lets consider each place on the number-plate. The first letter = 26 P 1. The second letter = 26 P 1. The third letter = 26 P 1. The fourth letter = 26 P 1. The first digit = 10 P 1 The second digit = 10 P 1. N o of car number-plates= 26 P 1 x 26 P 1 x 26 P 1 x 26 P 1 x 10 P 1 x 10 P 1 = 26 x 26 x 26 x 26 x 10 x 10 = 45697600 Permutations of n-elements chosen r times, with replacement, is given by: Arrangements = 26 4 x10 2 = 45697600

4. Example 2 Car number-plates are made with 4 upper case letters from the alphabet (26) followed by 2 digits from 0 to 9. How many number-plates start with the letter ‘Z’ and end with the digit ‘9’? letters digits In this case, letters and digit ‘9’ can be repeated: ZAAA 09ZBBA 99ZXPZ 29….. Lets consider each place on the number- plate, place the ‘Z’ and ‘9” first. The second letter = 26 P 1. The third letter = 26 P 1. The fourth letter = 26 P 1. The first digit = 10 P 1. N o of car number-plates= 1 P 1 x 26 P 1 x 26 P 1 x 26 P 1 x 10 P 1 x 1 P 1 = 1 x 26 x 26 x 26 x 10 x 1 = 175760 The first letter ‘Z’ = 1 P 1. Z The Last digit ’9’ = 1 P 1 9

5. Example 3 In how many ways can 5 identical Mathematics books, 4 identical Physics books and 3 identical Biology books be arranged on a shelf if: a) there is no restrictions? b) the books are to be grouped by subject? 9 Maths 8 Physics7 Biology8 Physics 7 Biology a) there is no restrictions? Repetition of Textbooks: Maths = 5 P 5 = 5! Physics = 4 P 4 = 4! Biology = 3 P 3 = 3! If all the books where different colours, the number of arrangements = 12 P 12 = 12! Arrangements on a shelf b) the books are to be grouped by subject If the books separated into subject, there are only 3 GROUPS. Subject arrangements on a shelf = 3 P 3 = 3! = 6

6. Example 4 How many arrangements can be made by the letters of the word “DEFINITION”, if a) there are no restrictions? b) the letter ‘I’ or ‘N’ do not occur in the first and last place? DEFIONTNII Repetition of LETTERS: ‘N’ = 2 P 2 = 2! ‘I’ = 3 P 3 = 3! a) there are no restrictions arrangements b) the letter ‘I’ or ‘N’ do not occur in the first and last place arrangements In this case, allocate the first and last place NOT using the ‘I’s or ‘N’s First and last place Now allocate the rest of the position using the 5 letters plus the 3 ‘I’s. Middle places

7. Example 5 The letters {A, E, I, O, U } are vowels. (a) How many arrangements of the letters in the word ‘ALGEBRAIC’ are possible? (b) How many arrangements of the letters in the word ‘ALGEBRAIC’ are possible if the vowels must occupy the 2 nd, 3 rd and 5 th position? CLGEBRAAI There are 2’A’= 2 P 2 (a) arrangements Place Vowels = {A, A, E, I} Consonants = {L, G, B, R, C} = 5 P 5 = 5! (b) arrangements =