PERMUTATIONS
- 2. This video is designed for grade 11 math students. For more videos, visit my page https://www.mathmadeeasy.co/lessons
- 3. This video /presentation explains • The fundamental principle of counting • The concept of permutations • Permutation formula • Calculating permutation for objects grouped together
- 4. Fundamental Principle Of Counting If an event A can occur in m ways, another event can occur in n ways, then both together can occur in mn ways.
- 5. A permutation is an arrangement which can be made by taking some or all objects together.
- 6. Number of permutations of n objects taken r at a time is denoted by )!( ! rn n pr n It is also denoted by P(n.r) n!=n(n-1)(n-2)(n-3)---------1
- 7. 1)How many 3 digit numbers can be formed using the digits 1 to 9 if i. no digit is repeated ii. digits are repeated If no digit is repeated, the 1st digit can be filled in 9 ways The 2nd digit in 8 ways The 3rd digit in 7 ways Answer = 9 × 8 × 7 = 504 𝑤𝑎𝑦𝑠
- 8. If digits are repeated, the 1st digit can be filled in 9 ways The 2nd digit in 9 ways The 3rd digit in 9 ways Ans = 93 = 729
- 9. 2) There are 8 true or false questions in a question paper. How many sequences of answers are possible? Each question can be answered in 3 ways, True, False, or Blank. Total no of ways = 3(8)= 24
- 10. 3) In how many ways can 8 books be arranged on a shelf if (i) any arrangement is possible (ii) 3 particular books must always stand together (iii) 2 particular books occupy the end
- 11. (i) If any arrangement is possible the 8 books can be arranged in 8!= 40320 ways. (ii) Treat the 3 books as one. There are now 8-3+1 = 6 books. These can be arranged in 6 ! Ways. The 3 books within themselves can be arranged in 3! ways. Ans = 6 !(3!) = 4320
- 12. (iii) Arrange the 2 particular books at the end. (1 __ __ __ __ __ __ __ 2 or 2 ___ __ __ __ __ __ __ 1) The remaining 6 books can be arranged in 6! ways . The 2 books within themselves can be arranged in 2 ways. Ans = 6! (2) = 1440
- 13. 4)In how many ways can 4 boys and 3 girls be seated in a row so that no 2 girls sit together. __ B__ B__ B__ B__ Arrange the 4 boys in a row. This can be done in 4! ways. Since no 2 girls sit together there are 5 spaces left for the girls.
- 14. These can be filled in 3 5 p ways Required answer = 4! ( 3 5 p ) = 1440 ways
- 15. 4)How many restrictions can be made from the letters of the word VOWELS if (i) There is no restriction. (ii) Each word begins with S and ends with E (iii) All the vowels come together.
- 16. (i) The word VOWELS has 6 letters none of which is repeated. The letters can be arranged in 6! ways when there is no restriction. (ii) S __ ___ __ __ E Fix the first and the last letter. The remaining 4 letters can be arranged in 4! ways.
- 17. (iii) The word VOWELS has 2 vowels O,E. Treat the 2 vowels as one. There are now 5 letters V,W,L,S +1. These can be permuted in 5! Ways. The 2 vowels O,E can be arranged in 2 ways. Total no of ways = 5!(2)= 240
- 18. I hope by the end of this video, you are able to understand the concept of • A permutation • How to evaluate permutations • How to evaluate permutations of grouped objects/letters with or without restriction
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