1) The document discusses permutations and combinations, providing examples of calculating the number of arrangements of letters in words. 2) It explains the formulas for permutations when objects are distinct or indistinct, and when objects occur together or separately. 3) Examples are given of calculating combinations by choosing objects without regard to order, such as choosing playing cards. 4) The key difference between permutations and combinations is explained - permutations consider order while combinations do not.

1. V.V. VANNIAPERUMAL COLLEGE FOR WOMEN
VALUE ADDED COURSE
Permutations and Combinations
Dr. T. Anitha
Assistant Professor
Department of Mathematics(SF)

2. How many ways can we arrange the letters of the word MAT?
MAT
MTA
ATM
AMT
TMA
TAM
3𝑃3 =
3!
3 − 3 !
=
3!
0!
= 3! = 3 × 2 × 1 = 6

3. Type 2: Objects are not distinct
How many ways can we arrange the letters of the word MATHEMATICS?
Total number of letters=11
Number of Ms= 2
Number of As = 2
Number of Ts = 2
Number of permutations =
11!
2!2!2!
=
11×10×9×8×7×6×5×4×3×2×1
2×2×2
= 4,989,600

4. Type 2: Objects are not distinct
Number of permutations of n objects where
1st object repeated 𝑝1 times
2nd object repeated 𝑝2 times
⋮
nth object repeated 𝑝𝑛 times
Then the permutation is =
𝑛!
𝑝1!𝑝2!⋯𝑝𝑛!

5. Type 3: Objects occur together
Find the number of different 8-letter arrangements that can be made from letters of the word
DAUGHTER so that
(i) All vowels occur together (ii) all vowels do not occur together
(i) D A U G H T E R
AUE D G H T R
6! ways arrange 6 letters and 3! ways arrange the vowels.
Required number of ways = 6! × 3!

6. (ii) The number of ways arrange the given 8 letters is 8!
Number of ways arrange the letters such that vowels occur together is
6! × 3!
Number of ways arrange the letters such that vowels occur together is
8! − (6! × 3!)

7. Try it yourself
In how many ways of the distinct permutations of the letters in
MISSISSIPPI do the four I’s not

8. COMBINATION

9. Selection and arranging are independent decisions, so
 Selection and Arrangement = Selection × Arrangement
 𝑛𝑃𝑟 = Selection × 𝑟!
 selection =
𝑛𝑃𝑟
𝑟!
 𝑛𝐶𝑟 =
𝑛!
𝑛−𝑟 !𝑟!

10. In how many ways you can choose 8 of 32 playing cards not considering
their order?
𝑛𝐶𝑟 = 32𝐶8 =
32!
24! 8!
=
32×31×⋯×25×24!
8×7×6×5×4×3×2×1×24!
= 10518300

11. PERMUTATION VS COMBINATION
PERMUTATION
 FORM 2 LETTER WORD FROM A 3
ALPHABETS {A,B,C}
 AB AC BC BA CA CB
 6 WAYS.
𝑛𝑃𝑟 = 3𝑃2 =
3!
3−2 !
= 6
FORM 3 LETTER WORD FROM A 3
ALPHABETS {A,B,C}
 ABC ACB BCA BAC CAB CBA
 6 WAYS.
𝑛𝑃𝑟 = 3𝑃3 =
3!
3−3 !
= 6
COMBINATION
SELECT 2 LETTERS FROM A 3 ALPHABETS
{A,B,C}
 AB AC BC
 3 WAYS.
𝑛𝐶𝑟 = 3𝐶2 =
3!
3−2 ! 2!
= 3
SELECT 3 LETTERS FROM A 3 ALPHABETS
{A,B,C}
 ABC
 1 WAYS.
𝑛𝐶𝑟 = 3𝐶3 =
3!
3−3 ! 3!
= 1

12. THAN YOU