What are permutations? where is it used? This video/presentation introduces permutations and how to calculate it. It shows how permutations can be evaluated when the objects are arranged together or separately. It also explains permutations with or without repetition. This is useful for students of grade 11 mathematics, GRE QUANT maths and maths for the NATA Eligibility test in Architecture. For more videos visit my page https://www.mathmadeeasy.co/lessons

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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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